11. 盛最多水的容器
- 2019 年 10 月 4 日
- 筆記
題目描述
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
符合直覺的解法是，我們可以對兩兩進行求解，計算可以承載的水量。然後不斷更新最大值，最後返回最大值即可。這種解法，需要兩層循環，時間複雜度是O(n^2)
eg:
// 這個解法比較暴力，效率比較低 // 時間複雜度是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;
這種符合直覺的解法有點像冒泡排序， 大家可以稍微類比一下
那麼有沒有更加優的解法呢？我們來換個角度來思考這個問題，上述的解法是通過兩兩組合，這無疑是完備的， 那我門是否可以先計算長度為n的面積，然後計算長度為n-1的面積，… 計算長度為1的面積。這樣去不斷更新最大值呢？很顯然這種解法也是完備的，但是似乎時間複雜度還是O(n ^ 2), 不要着急。
考慮一下，如果我們計算n-1長度的面積的時候,是直接直接排除一半的結果的。
如圖：
比如我們計算n面積的時候，假如左側的線段高度比右側的高度低，那麼我們通過左移右指針來將長度縮短為n-1的做法是沒有意義的， 因為 新的形成的面積變成了(n-1)*heightOfLeft這個面積一定比剛才的長度為n的面積nn*heightOfLeft小
也就是說最大面積 一定是當前的面積或者通過移動短的線段得到。
關鍵點解析
- 雙指針優化時間複雜度
代碼
- 語言支持：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; // 雙指針來進行優化 // 時間複雜度是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 { // 如果相等就隨便了 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; } };
