Leetcode #11. Container With Most Water

原题：

Given n non-negative integersa1, a2, ...,an, where each represents a pointat 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.

题型：辅助变量；数组；指针。

解法：

1.暴力遍历法，时间为O(n^2)。

2.借助两个指针，首部指针l和尾部指针r。从两端向中间遍历：每一次都算出选取当前l和r值时的面积，若比ans大，则替换之；然后比较r和l的值谁的小，较小的那个指针向中间挪一位。利用木桶原理：由于面积被较小边限制，所以只有当较小的一方向中间挪一位才有可能得到更大的面积，这是一步相当有效的剪枝。

代码：

class Solution {public:    int maxArea(vector<int>& height) {        int l = 0, r = height.size() - 1;        int ans = 0;        while(l != r)        {            int size;            if(height[l] >= height[r])            {                size = (r - l) * height[r];                r --;            }            else            {                size = (r - l) * height[l];                l ++;            }            ans = ans < size ? size : ans;        }        return ans;    }};