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。
         * */
    }
};