标签：
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.
思路：
肯定是要确定水槽的左右边界位置，需要两个变量l1,l2 分别从最左边和最右边收缩。收缩规则是，比较矮的那一边向中间靠拢。更新水量。
我自己的代码：
int maxArea(vector<int> &height) { int ans = (height.size() - 1) * min(height[0], height.back()) ; int l1, l2, h1, h2; h1 = 0; h2 = height.size() - 1; l1 = h1; l2 = h2; while(l1 < l2) { if(height[l1] > height[h1]) { h1 = l1; int tmp = (l2 - l1) * min(height[l2], height[l1]); ans = (tmp > ans) ? tmp : ans; } if(height[l2] > height[h2]) { h2 = l2; int tmp = (l2 - l1) * min(height[l2], height[l1]); ans = (tmp > ans) ? tmp : ans; } if(height[l1] < height[l2]) { l1++; } else if(height[l1] > height[l2]) { l2--; } else { l1++; l2--; } } return ans; }
大神的代码就简洁很多：
int maxArea(vector<int> &height) { int left = 0, right = height.size() - 1; int maxWater = 0; while(left < right){ maxWater = max(maxWater, (right - left) * min(height[left], height[right])); height[left] < height[right] ? left++ : right--; } return maxWater; }
【leetcode】Container With Most Water（middle）
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原文地址：http://www.cnblogs.com/dplearning/p/4275912.html
