LeetCode 之 Container With Most Water

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.

首先要理解题的意思，就是找到容积最大的桶，假设有两个指针（i,j）height[i]和height[j]中的最小者与i和j的差值决定了桶的容积，从这里我们可以想到短板效应。思路如下：两个指针刚开始一个在头一个在尾，之后向中间走，由于桶的底在变短，所以要获得最大的容积必须要两个边的最小值大于之前的最小值。在写代码时思路就是让i,j中小的移动，让短板变大。代码如下：

class Solution {public:    int maxArea(vector<int>& height) {        int max_area=0;        int length=height.size();                int i=0,j=length-1;        while(i<j){            int temp_max=(j-i)*(min(height[i],height[j]));            max_area=max(temp_max,max_area);            if(height[i]<height[j]){                i++;            }else{                j--;            }        }        return max_area;    }};