【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.

解析：给一组数组作为水池的高度，以其中两个高度作为水池的壁，求最大的水池容量。假设左边的壁为left，右边的壁为right，则水池容量：

capacity= (right-left)*min(height(right),height(right))

解决方法是定义两个指针left和right从数组两头开始遍历：如果left的高度低于right，则将left指针右移找到比left高的重新判定capacity；反之，如果right的高度低于left，则将right的指针左移找到比right高的重新判定capacity。最后left==right结束while循环。

C++ AC代码：

class Solution {public:    int maxArea(vector<int> &height) {        int capacity = 0;        int left = 0,right = height.size()-1;        int k;        while(left<right){            capacity = max(capacity,(right-left)*min(height[left],height[right]));                       if(height[left]<height[right]){                k = left;                while(k<right && height[k] <= height[left]){                    k++;                }                left = k;            }else{                k = right;                while(k>left && height[k] <= height[right]){                    k--;                }                right = k;            }        }        return capacity;    }};