Container With Most Water —— Leetcode

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.

刚开始想到的是动态规划问题，但是，没有重叠的子问题，这显然不是动态规划，看下面简洁的代码：

class Solution {public:    int maxArea(vector<int>& height) {        /**        container的容量遵循以下事实；        1. 最宽的肯定是第一候选人；        2. 如果不宽，那么我们就需要更高的water level；        3. 低的water level我们可以直接删掉        */                int i = 0, j = height.size()-1;        int water = 0;        while(i < j) {            water = max(water, (j-i)*min(height[i], height[j]));            if(height[i] < height[j])                i ++;            else                j --;        }        return water;                /**         * 该代码可以改进的地方是，把小于容器中间的所有lower level全部排除，速度会更快。         * */    }};