[LeetCode]Container With Most Water

题目：
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.

分析：
面积 = 底 * 高 =（right - left） * min(a[left ] , a[right ]) （此处right > left）

首先让底最大即left = 0,right = len - 1，把计算出来的面积的保存在ans里面，ans始终用来存储已经计算出来的面积的最大值，然后不断地把底缩小，找更大的面积。

若a[left] < a[right]:
说明要找更大的面积，对于底不变，我们现在必须要找高更大的值，也就是两个高中较小的那个值必须去找更大的值，才能使面积变大
所以此时 left ++

反之同理

class Solution {public:    int maxArea(vector<int>& height) {        int len = height.size();        if(len < 2)            return false;        int ans = 0;    //用来记录已经算出的最大的面积        //面积 = 底 * 高=(right - left)* min(a[i],a[j])        int left = 0;        int right = len - 1;        while(left < right)        {            ans = max(ans, (right - left) * min(height[left], height[right]));            if(height[left] < height[right])            {                left++;            }            else            {                right--;            }        }        return ans;    }};