leetcode: 11. Container With Most Water

题意：

输入n个数，每个数的值a[i]为该数的高（a[i] >= 0），其位置为该数的输入顺序（间隔为1），要求从这堆输入的数中找到两个高，使得这两个高之间的容器能装最多的水，注意：容器不能倾斜。

解题思路：

从离得最远的两个数入手，最远的两数拥有最大的宽度，然后当前的高为头尾两数的最小值，用宽高求出当前水容量ans。接着先从左往右寻找第一个比当前高要高的左下标i，再从右往左找比当前高的右下标j，再求当前两下标的水容量与ans相比，取最大者作为新的ans，直至i>=j时结束算法。其时间复杂度为O(n)。

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



                                             
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