【Leetcode 做题学算法周刊】第三期

  • 2019 年 11 月 13 日
  • 笔记

首发于微信公众号《前端成长记》，写于 2019.11.13

背景

本文记录刷题过程中的整个思考过程，以供参考。主要内容涵盖：

  • 题目分析设想
  • 编写代码验证
  • 查阅他人解法
  • 思考总结

目录

Easy

35.搜索插入位置

题目地址

题目描述

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

你可以假设数组中无重复元素。

示例：

输入: [1,3,5,6], 5  输出: 2    输入: [1,3,5,6], 2  输出: 1    输入: [1,3,5,6], 7  输出: 4    输入: [1,3,5,6], 0  输出: 0

题目分析设想

这道题目有点明显，题干说明了是排序数组，重点是排序数组，所以很明显的第一反应会使用二分法来解题。同时可以注意一下，数组中无重复元素。所以这道题我就按两个方案来作答：

  • 暴力法，直接遍历
  • 二分法，可以理解成不断折半排除不可能

编写代码验证

Ⅰ.暴力法

代码：

/**   * @param {number[]} nums   * @param {number} target   * @return {number}   */  var searchInsert = function(nums, target) {      if (nums.length === 0 || nums[0] > target) return 0;      if(nums[nums.length - 1] < target) return nums.length;        for(let i = 0; i < nums.length; i++) {          if(nums[i] >= target) return i      }  };

结果：

  • 62/62 cases passed (60 ms)
  • Your runtime beats 92.48 % of javascript submissions
  • Your memory usage beats 63.22 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

Ⅱ.二分法

代码：

/**   * @param {number[]} nums   * @param {number} target   * @return {number}   */  var searchInsert = function(nums, target) {      if (nums.length === 0 || nums[0] > target) return 0;      if(nums[nums.length - 1] < target) return nums.length;        let left = 0; // 起点      let right = nums.length - 1; // 终点      while(left < right) {           // 零填充右位移，使用位运算避免溢出，大部分情况下等于 (left + right / 2)          let i = parseInt((left + right) >>> 1) // 这里选择取左          if (nums[i] < target) { // 中位数小于目标值              left = i + 1 // 排除中位数左侧          } else {              right = i // 排除中位数右侧          }      }      return left  };

结果：

  • 62/62 cases passed (52 ms)
  • Your runtime beats 99.31 % of javascript submissions
  • Your memory usage beats 61.31 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(log2(n))

查阅他人解法

基本上这道题就是针对二分法进行考察的，所以没有看到其他特别的解法。

思考总结

看见排序数组，查找下标，那么就可以果断选择二分法啦。

38.报数

题目地址

题目描述

报数序列是一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：

1.     1  2.     11  3.     21  4.     1211  5.     111221

1 被读作 "one 1" ("一个一"),即 11。

11 被读作 "two 1s" ("两个一"),即 21。

21 被读作 "one 2", "one 1" （"一个二" , "一个一") , 即 1211。

给定一个正整数 n（1 ≤ n ≤ 30），输出报数序列的第 n 项。

注意：整数顺序将表示为一个字符串。

示例:

输入: 1  输出: "1"    输入: 4  输出: "1211"

题目分析设想

这道题目有点绕，其实我们只看右侧项就可以，每次都是读上一次项。报数的规则实际上就是相同连续数字合并后进行每位的报数。最简单的想法是直接使用正则替换就可以了。当然也可以从递归和遍历的方式来作答，我们分别来看看。

  • 正则法，替换相同连续数字为 长度 + 数字本身
  • 递归法，不断转换成 n-1 求解
  • 遍历法，不断转换成 n+1 求解

编写代码验证

Ⅰ.正则法

代码：

/**   * @param {number} n   * @return {string}   */  var countAndSay = function(n) {      let str = '1'      for(let i = 1; i < n; i++) {          // 匹配项长度+第一位即为读数          str = str.replace(/(d)1*/g, (match) => (`${match.length}${match.charAt(0)}`))      }      return str  };

结果：

  • 18/18 cases passed (56 ms)
  • Your runtime beats 98.81 % of javascript submissions
  • Your memory usage beats 32.53 % of javascript submissions (35.6 MB)
  • 时间复杂度 O(n)

Ⅱ.递归法

代码：

/**   * @param {number} n   * @return {string}   */  var countAndSay = function(n) {      if (n === 1) return '1'      debugger      let str = countAndSay(n - 1)      let item = str.charAt(0)        let count = 0 // 计数器      let res = ''      for(let i = 0; i < str.length; i++) {          if(str.charAt(i) === item) {              count++          } else {              res += `${count}${item}`              item = str.charAt(i)              count = 1          }            if (i === str.length - 1) { // 最后一位，需要取数              res += `${count}${item}`          }      }      return res  };

结果：

  • 18/18 cases passed (64 ms)
  • Your runtime beats 92.23 % of javascript submissions
  • Your memory usage beats 28.23 % of javascript submissions (35.7 MB)
  • 时间复杂度 O(n^2)

Ⅲ.遍历法

代码：

/**   * @param {number} n   * @return {string}   */  var countAndSay = function(n) {      let str = '1'      for(let i = 1; i < n; i++) {          str = countEach(str)      }      function countEach(str) {          let count = 0          let res = ''          for(let i = 0; i < str.length; i++) {              if(i === 0 || str.charAt(i) === str.charAt(i - 1)) {                  count++              } else {                  res += `${count}${str.charAt(i - 1)}`                  count = 1              }              if (i === str.length - 1) {                  res += `${count}${str.charAt(i)}`              }          }            return res      }        return str  };

结果：

  • 18/18 cases passed (60 ms)
  • Your runtime beats 96.98 % of javascript submissions
  • Your memory usage beats 41.69 % of javascript submissions (35.5 MB)
  • 时间复杂度 O(n^2)

查阅他人解法

这里看到一个开怀一笑的解法，直接字典法，缺点很明显，但是当前情况下确实是最快的。

Ⅰ.正则法

代码：

/**   * @param {number} n   * @return {string}   */  var countAndSay = function(n) {      const map = {          1:"1",          2:"11",          3:"21",          4:"1211",          5:"111221",          6:"312211",          7:"13112221",          8:"1113213211",          9:"31131211131221",          10:"13211311123113112211",          11:"11131221133112132113212221",          12:"3113112221232112111312211312113211",          13:"1321132132111213122112311311222113111221131221",          14:"11131221131211131231121113112221121321132132211331222113112211",          15:"311311222113111231131112132112311321322112111312211312111322212311322113212221",          16:"132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211",          17:"11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221",          18:"31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211",          19:"1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221",          20:"11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211",          21:"311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311122122111312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221131112311311121321122112132231121113122113322113111221131221",          22:"132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113213221133122112231131122211211131221131112311332211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211",          23:"111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121113222123112221221321132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221",          24:"3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211",          25:"132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113212231121113112221121321132132211231232112311321322112311311222113111231133221121113122113121113221112131221123113111231121123222112132113213221133112132123123112111312111312212231131122211311123113322112111312211312111322111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312212221121123222112132113213221133112132123123112111311222112132113213221132213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121132211332113221122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321231231121113112221121321132122311211131122211211131221131211322113322112111312211322132113213221123113112221131112311311121321122112132231121113122113322113111221131221",          26:"1113122113121113222123211211131211121311121321123113213221121113122123211211131221121311121312211213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121113222112131112131221121321131211132221121321132132211331121321232221123113112221131112311322311211131122211213211331121321122112133221121113122113121113222123211211131211121311121321123113111231131122112213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122113221122112133221121113122113121113222123211211131211121311121321123113213221121113122113121113222113221113122113121113222112132113213221232112111312111213322112311311222113111221221113122112132113121113222112311311222113111221132221231221132221222112112322211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112211322212322211231131122211322111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211",          27:"31131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322123211211131211121332211231131122211311122122111312211213211312111322211231131122211311123113322112111331121113112221121113122113111231133221121113122113121113222123211211131211121332211213211321322113311213211322132112311321322112111312212321121113122122211211232221123113112221131112311332111213122112311311123112111331121113122112132113311213211321222122111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311123113322113223113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331221122311311222112111312211311123113322112132113213221133122211332111213112221133211322112211213322112111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121122132112311321322112111312211312111322211213111213122112132113121113222112132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212311222122132113213221123113112221133112132123222112111312211312111322212321121113121112133221121311121312211213211312111322211213211321322123211211131211121332211213211321322113311213212312311211131122211213211331121321122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311222113111221221113122112132113121113222112132113213221133122211332111213322112132113213221132231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221",          28:"13211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221232112111312211312113211223113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321322113311213212322211322132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212311222122132113213221123113112221133112132123222112111312211312111322212311322123123112111321322123122113222122211211232221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112311311222113111231133211121321321122111312211312111322211213211321322123211211131211121332211231131122211311123113321112131221123113111231121123222112111331121113112221121113122113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132132211331221122311311222112111312211311123113322112111312211312111322212311322123123112112322211211131221131211132221132213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211",          29:"11131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221231122212213211321322112311311222113311213212322211211131221131211132221232112111312111213322112131112131221121321131211132221121321132132212321121113121112133221121321132132211331121321231231121113112221121321133112132112211213322112311311222113111231133211121312211231131122211322311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122111213122112311311222113111221131221221321132132211331121321231231121113112221121321133112132112211213322112311311222113111231133211121312211231131122211322311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122111213122112311311222112111331121113112221121113122113121113222112132113213221232112111312111213322112311311222113111221221113122112132113121113222112311311222113111221132221231221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312111322212321121113121112133221132211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112132132112211131221131211132221121321132132212321121113121112133221123113112221131112311332111213211322111213111213211231131211132211121311222113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131211131221223113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112211213322112312321123113213221123113112221133112132123222112311311222113111231132231121113112221121321133112132112211213322112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122111213122112311311221132211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312111322212311222122132113213221123113112221133112132123222112311311222113111231133211121321132211121311121321122112133221123113112221131112311332211322111312211312111322212321121113121112133221121321132132211331121321231231121113112221121321132122311211131122211211131221131211322113322112111312211322132113213221123113112221131112311311121321122112132231121113122113322113111221131221",          30:"3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112311311222113111231133211121321321122111312211312111322211213211321322123211211131211121332211231131122211311123113321112131221123113111231121123222112111331121113112221121113122113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321223112111311222112132113213221133122211311221122111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331222113321112131122211332113221122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112112322211322311311222113111231133211121312211231131112311211232221121113122113121113222123211211131221132211131221121321131211132221123113112211121312211231131122113221122112133221121321132132211331121321231231121113121113122122311311222113111231133221121113122113121113221112131221123113111231121123222112132113213221133112132123123112111312211322311211133112111312211213211311123113223112111321322123122113222122211211232221121113122113121113222123211211131211121311121321123113213221121113122123211211131221121311121312211213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121113222112131112131221121321131211132221121321132132211331121321232221123113112221131112311322311211131122211213211331121321122112133221121113122113121113222123112221221321132132211231131122211331121321232221121113122113121113222123211211131211121332211213111213122112132113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231131112311311221122132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121311121312211213211312111322211213211321322123211211131211121332211213211321322113311213211322132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132122311211131122211213211321222113222122211211232221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111312211322311211133112111312212221121123222112132113213221133112132123222113223113112221131112311332111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112211322212322211231131122211322111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211"      }        return map[n]  };

结果：

  • 18/18 cases passed (56 ms)
  • Your runtime beats 98.81 % of javascript submissions
  • Your memory usage beats 98.5 % of javascript submissions (33.5 MB)
  • 时间复杂度 O(1)

思考总结

总体而言，这道题用正则去解答十分简单，考验的点在正则的匹配这块；当然递归或者遍历也是常规思路；字典法纯属一乐。推荐使用正则来解答此题。

53.最大子序和

题目地址

题目描述

给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

示例:

输入: [-2,1,-3,4,-1,2,1,-5,4],  输出: 6  解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。

进阶:

如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。

题目分析设想

这道题首先基本解法肯定是暴力的遍历求解，直接遍历找出最大区间。当然这里我们也可以使用动态规划来思考问题，列出动态和转移方程式，等于求解 Max(d[0, i])。另外进阶里面提示分治法，分治法在之前有用过，我们也可以做为一个方向。所以大概有三种：

  • 遍历求解，直接遍历算出各区间值
  • 动态规划问题，求解动态问题，找到每个动态区间的最大值
  • 分治，不断二分找区间内的最大子序和

注意一下，只要是寻找最大值最小值的，初始值需要定义为理论上的最大最小值。

编写代码验证

Ⅰ.遍历求解

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      let res = Number.MIN_SAFE_INTEGER      for(let i = 0; i < nums.length; i++) {          let sum = 0          // 分别算出 i 开始的最大子序和          for(let j = i; j < nums.length; j++) {              sum += nums[j];              res = Math.max(res, sum)          }      }      return res  };

结果：

  • 202/202 cases passed (272 ms)
  • Your runtime beats 7.61 % of javascript submissions
  • Your memory usage beats 41.88 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n^2)

Ⅱ.动态规划

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      let res = dp = nums[0] // 初始值      for(let i = 1; i < nums.length; i++) {          dp = Math.max(dp + nums[i], nums[i]) // 动态取得最大值          res = Math.max(res, dp)      }      return res  };

结果：

  • 202/202 cases passed (68 ms)
  • Your runtime beats 90.14 % of javascript submissions
  • Your memory usage beats 45 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅲ.分治

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      // 不断分治      function countByDichotomy (start, end) {          // 存储左侧结果，右侧结果，两者更大值，以及两者相加的值          // 解释一下：最大子序列在左右两区间内要么过界要么不过界。          // 如果不过界，则最大值为 Max(left, right)          // 如果过界，则最大为左区间到中间的最大值加中间到右区间的最大值          if (end === start) { // 数组就一项              return {                  lmax: nums[start], // 左半区间包含其右端点的最大子序和                  rmax: nums[start], // 右半区间包含其左端点的最大子序和                  sum: nums[start], // 总和                  result: nums[start] // 区域内部的最大子序和              }          } else {              const mid = (start + end) >>> 1 // 这个取中位数写法之前解释过，避免溢出              const left = countByDichotomy(start, mid) // 左区间中计算结果              const right = countByDichotomy(mid + 1, end) // 右区间中计算结果              return {                  lmax: Math.max(left.lmax, left.sum + right.lmax),                  rmax: Math.max(right.rmax, left.rmax + right.sum),                  sum: left.sum + right.sum,                  result: Math.max(left.rmax + right.lmax, Math.max(left.result, right.result))              }          }      }      return countByDichotomy(0, nums.length - 1).result;  };

结果：

  • 202/202 cases passed (60 ms)
  • Your runtime beats 97.89 % of javascript submissions
  • Your memory usage beats 5.01 % of javascript submissions (36.7 MB)
  • 时间复杂度 O(n)

查阅他人解法

查阅题解的过程中发现了以下几种有意思的思路：

  • 动态规划，使用增益的思路。其实上面我们写的动态规划是一样的
  • 贪心法，尝试多加一位，取较大值
  • 分治中使用贪心法求区间

Ⅰ.动态规划

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      let res = nums[0]      let sum = nums[0] // 增益      for(let i = 1; i < nums.length; i++) {          if (sum > 0) { // 正向增益， sum 保留并加上当前遍历数字              sum += nums[i]          } else { // sum 更新为当前遍历数字              sum = nums[i]          }          res = Math.max(res, sum)      }      return res  };

结果：

  • 202/202 cases passed (60 ms)
  • Your runtime beats 97.89 % of javascript submissions
  • Your memory usage beats 47.5 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅱ.贪心法

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      let res = Number.MIN_SAFE_INTEGER // 初始值      let sum = 0      for(let i = 0; i < nums.length; i++) {          sum += nums[i]          res = Math.max(res, sum)          if (sum < 0) { // 重新开始找子序串              sum = 0;          }      }      return res  };

结果：

  • 202/202 cases passed (68 ms)
  • Your runtime beats 90.14 % of javascript submissions
  • Your memory usage beats 45 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅲ.分治中使用贪心法求区间

代码：

/**   * @param {number[]} nums   * @return {number}   */  var maxSubArray = function(nums) {      // 获取跨边界的和      function getMaxCross (start, mid, end) {          let leftRes = Number.MIN_SAFE_INTEGER          let leftSum = 0          for(let i = mid; i >= start; i--) {              leftSum += nums[i]              leftRes = Math.max(leftRes, leftSum)          }            let rightRes = Number.MIN_SAFE_INTEGER          let rightSum = 0          for(let i = mid + 1; i <= end; i++) {              rightSum += nums[i]              rightRes = Math.max(rightRes, rightSum)          }            return leftRes + rightRes      }        function countByDichotomy (start, end) {          if (start === end) {              return nums[start]          } else {              const mid = (start + end) >>> 1              const leftSum = countByDichotomy(start, mid)              const rightSum = countByDichotomy(mid + 1, end)              const midSum = getMaxCross(start, mid, end)              // 三者比较最大的就为最大子序和              return Math.max(leftSum, rightSum, midSum)          }      }        return countByDichotomy(0, nums.length - 1)  };

结果：

  • 202/202 cases passed (72 ms)
  • Your runtime beats 80.56 % of javascript submissions
  • Your memory usage beats 49.38 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(nlog(n))

思考总结

个人认为动态规划在这套题里面解题思路清晰，贪心法也可以理解为基于遍历基础上做的延伸，而分治法需要画图加以理解。一般看到这种最大最长的题目，基本上就可以用动态规划问题来尝试作答了。

58.最后一个单词的长度

题目地址

题目描述

给定一个仅包含大小写字母和空格 ' ' 的字符串，返回其最后一个单词的长度。

如果不存在最后一个单词，请返回 0 。

说明：一个单词是指由字母组成，但不包含任何空格的字符串。

示例：

输入: "Hello World"  输出: 5

题目分析设想

这道题看上去像是一道字符串题，我们可以从以下几个方面来尝试作答：

  • 遍历，从末尾开始，效率高
  • lastIndexOf，直接找空格
  • 正则
  • split

编写代码验证

Ⅰ.遍历

代码：

/**   * @param {string} s   * @return {number}   */  var lengthOfLastWord = function(s) {      if(!s.length) return 0      let i = s.length - 1      while(i >= 0 && s.charAt(i) === ' ') {          i--      }      if(i < 0) return 0 // 全是空格        let j = i      while(j >= 0 && s.charAt(j) != ' ') {          j--      }      return i - j  };

结果：

  • 59/59 cases passed (64 ms)
  • Your runtime beats 81.14 % of javascript submissions
  • Your memory usage beats 29.72 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

Ⅱ.lastIndexOf

代码：

/**   * @param {string} s   * @return {number}   */  var lengthOfLastWord = function(s) {      if(!s.length) return 0      s = s.trim()      const idx = s.lastIndexOf(' ')      return idx === -1 ? s.length : s.length - 1 - idx  };

结果：

  • 59/59 cases passed (48 ms)
  • Your runtime beats 99.48 % of javascript submissions
  • Your memory usage beats 36.52 % of javascript submissions (33.7 MB)
  • 时间复杂度 O(1)

Ⅲ.正则

代码：

/**   * @param {string} s   * @return {number}   */  var lengthOfLastWord = function(s) {      if(!s.length) return 0      const match = s.match(/([a-zA-Z]+)s*$/)      let res = 0      if (match) {          res = match.pop()          return res.length      }      return res  };

结果：

  • 59/59 cases passed (80 ms)
  • Your runtime beats 26.65 % of javascript submissions
  • Your memory usage beats 5.95 % of javascript submissions (34.2 MB)
  • 时间复杂度 O(1)

Ⅳ.split

代码：

/**   * @param {string} s   * @return {number}   */  var lengthOfLastWord = function(s) {      if(!s.length) return 0      s = s.trim()      const arr = s.split(' ')      if (arr.length) {          let str = arr.pop()          return str.length      } else {          return 0      }  };

结果：

  • 59/59 cases passed (60 ms)
  • Your runtime beats 90.57 % of javascript submissions
  • Your memory usage beats 13.8 % of javascript submissions (34 MB)
  • 时间复杂度 O(1)

查阅他人解法

没有在题解中看到什么特别的解法，大部分都是基于类库解的，比如 Javascript 中 String 和 Array 的方法。或者是遍历实现的。

思考总结

直到现在也没有弄明白这道题的考察点在哪里？不过我建议感兴趣的同学，可以自己拓展实现 lastIndexOf ，参考 上一期 28题的数十种解法，应该能有不小收获。

66.加一

题目地址

题目描述

给定一个由整数组成的非空数组所表示的非负整数，在该数的基础上加一。

最高位数字存放在数组的首位， 数组中每个元素只存储单个数字。

你可以假设除了整数 0 之外，这个整数不会以零开头。

示例：

输入: [1,2,3]  输出: [1,2,4]  解释: 输入数组表示数字 123。    输入: [4,3,2,1]  输出: [4,3,2,2]  解释: 输入数组表示数字 4321。

题目分析设想

这道题我有两个大方向，一是数组遍历进行求解，另外一种是数组转数字再处理。但是转数字可能会溢出，所以就只想到从遍历的角度来作答。

  • 遍历，从后往前遍历找到不为9的项，后面填0就可以了

编写代码验证

Ⅰ.遍历

代码：

/**   * @param {number[]} digits   * @return {number[]}   */  var plusOne = function(digits) {      for(let i = digits.length - 1; i >= 0; i--) {          // 找不到不为9的数，直接加1输出就可以了          if(digits[i] !== 9) {              digits[i]++              return digits          } else {              digits[i] = 0          }      }      digits.unshift(1)      return digits  };

结果：

  • 109/109 cases passed (60 ms)
  • Your runtime beats 93.72 % of javascript submissions
  • Your memory usage beats 26.35 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

查阅他人解法

发现思路基本都是遍历，但是具体实现会有些差距，这里只列举一个最简单的。

Ⅰ.遍历

代码：

/**   * @param {number[]} digits   * @return {number[]}   */  var plusOne = function(digits) {      for(let i = digits.length - 1; i >= 0; i--) {          digits[i]++          // 取10的余数，做了赋值操作，为0就继续进位          digits[i] %= 10          if(digits[i] !== 0) {              return digits          }      }      digits.unshift(1)      return digits  };

结果：

  • 109/109 cases passed (64 ms)
  • Your runtime beats 85.29 % of javascript submissions
  • Your memory usage beats 94.34 % of javascript submissions (33.4 MB)
  • 时间复杂度 O(n)

思考总结

这道题可能大众的想法会转成数字再处理，但是做数字运算的时候，千万要记住考虑溢出的问题。另外因为是加1，所以倒序遍历就可以了。至于是判断末位为9还是对10取余，我觉得都是一个很好理解的思路，也避免了代码的繁琐。

（完）

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