class Solution { public: int maxArea(vector<int>& height) { ; ; ; while(l < r) { int h = min(height[l], height[r]); int area = h * (r-l); if (area > max) max = area; while (height[l] <= h && l < r) l++; while (height[r] <=…

Description 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 for…

我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ 011. Container With Most Water[M] 问题 Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai).…

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 containe…

class Solution { public: string longestPalindrome(string s) { int length = s.length(); ) return s; ; ; int **pS = new int *[length];//palindromic state ;i < length; i++) pS[i] = ]; ; i < length;i++) { pS[i][] = ; pS[i][] = ; } ; i <= length; i++)…

11. Container With Most Water https://www.cnblogs.com/grandyang/p/4455109.html 用双指针向中间滑动,较小的高度就作为当前情况的高度,然后循环找容量的最大值. 不管两个指针中间有多少高度的柱子,只管两头,因为两头的才决定最大容量. class Solution { public: int maxArea(vector<int>& height) { if(height.empty()) ; ; ,end = h…

11. Container With Most Water Medium 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…

一.题目说明 11.Container With Most Water,这个题目难度是Medium. 二.我的做法 乍一看,简单啊,两个for循环就可以了,我在本地写的. #include<iostream> #include<vector> #include<math.h> using namespace std; class Solution{ public: int maxArea(vector<int> &height){ int max =…

炎炎夏日,还是呆在空调房里切切题吧. Container With Most Water,题意其实有点噱头,简化下就是,给一个数组,恩,就叫 height 吧,从中任选两项 i 和 j(i <= j),使得 Math.min(height[i], height[j]) * (j - i) 最大化,求解这个最大值. O(n^2) O(n^2) 复杂度的解法非常容易想到,直接两两枚举. var maxArea = function(height) { var len = height.length;…

description: Input: [1,8,6,2,5,4,8,3,7]Output: 49 思路1: 从(1,a1)开始向后算面积,需要两层n循环,时间复杂度n2 思路2: 找出数组中最大的数,将其与次大,第三大数求面积,依次类推,也需要两层循环,还需要额外排序,时间复杂度n2 因为找出最大数并且并不知道输入数据的规律(有可能很杂乱),所以每个都有必要算,采取思路1. 代码实现如下: class Solution { public int maxArea(int[] height) {…