11. 盛最多水的容器

  • 2019 年 10 月 4 日
  • 笔记

题目描述

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.    Note: You may not slant the container and n is at least 2.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.        Example:    Input: [1,8,6,2,5,4,8,3,7]  Output: 49

思路

符合直觉的解法是,我们可以对两两进行求解,计算可以承载的水量。然后不断更新最大值,最后返回最大值即可。这种解法,需要两层循环,时间复杂度是O(n^2)

eg:

// 这个解法比较暴力,效率比较低      // 时间复杂度是O(n^2)      let max = 0;      for(let i = 0; i < height.length; i++) {          for(let j = i + 1; j < height.length; j++) {              const currentArea = Math.abs(i - j) * Math.min(height[i], height[j]);              if (currentArea > max) {                  max = currentArea;              }          }      }      return max;

这种符合直觉的解法有点像冒泡排序, 大家可以稍微类比一下

那么有没有更加优的解法呢?我们来换个角度来思考这个问题,上述的解法是通过两两组合,这无疑是完备的, 那我门是否可以先计算长度为n的面积,然后计算长度为n-1的面积,… 计算长度为1的面积。这样去不断更新最大值呢?很显然这种解法也是完备的,但是似乎时间复杂度还是O(n ^ 2), 不要着急。

考虑一下,如果我们计算n-1长度的面积的时候,是直接直接排除一半的结果的。

如图:

比如我们计算n面积的时候,假如左侧的线段高度比右侧的高度低,那么我们通过左移右指针来将长度缩短为n-1的做法是没有意义的, 因为 新的形成的面积变成了(n-1)*heightOfLeft这个面积一定比刚才的长度为n的面积nn*heightOfLeft小

也就是说最大面积 一定是当前的面积或者通过移动短的线段得到

关键点解析

  • 双指针优化时间复杂度

代码

  • 语言支持:JS,C++

JavaScript Code:

/*   * @lc app=leetcode id=11 lang=javascript   *   * [11] Container With Most Water   *   * https://leetcode.com/problems/container-with-most-water/description/   *   * algorithms   * Medium (42.86%)   * Total Accepted:    344.3K   * Total Submissions: 790.1K   * Testcase Example:  '[1,8,6,2,5,4,8,3,7]'   *   * Given n non-negative integers a1, a2, ..., an , where each represents a   * point at coordinate (i, ai). n vertical lines are drawn such that the two   * endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together   * with x-axis forms a container, such that the container contains the most   * water.   *   * Note: You may not slant the container and n is at least 2.   *   *   *   *   *   * The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In   * this case, the max area of water (blue section) the container can contain is   * 49.   *   *   *   * Example:   *   *   * Input: [1,8,6,2,5,4,8,3,7]   * Output: 49   *   */  /**   * @param {number[]} height   * @return {number}   */  var maxArea = function(height) {      if (!height || height.length <= 1) return 0;        // 双指针来进行优化      // 时间复杂度是O(n)      let leftPos = 0;      let rightPos = height.length - 1;      let max = 0;      while(leftPos < rightPos) {            const currentArea = Math.abs(leftPos - rightPos) * Math.min(height[leftPos] , height[rightPos]);          if (currentArea > max) {              max = currentArea;          }          // 更新小的          if (height[leftPos] < height[rightPos]) {              leftPos++;          } else { // 如果相等就随便了              rightPos--;          }      }        return max;  };

C++ Code:

class Solution {  public:      int maxArea(vector<int>& height) {          auto ret = 0ul, leftPos = 0ul, rightPos = height.size() - 1;          while( leftPos < rightPos)          {              ret = std::max(ret, std::min(height[leftPos], height[rightPos]) * (rightPos - leftPos));              if (height[leftPos] < height[rightPos]) ++leftPos;              else --rightPos;          }          return ret;      }  };