Container With Most Water

一、问题描述

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.
二、思路

本题难点在于理解题目。在二维坐标系中，求(i,0)(j,0)组成的线段乘以高度h组成的图形面积最大，是一道动态规划题目，我们可以采用双层for求解，但是那样显然不是最优的，我们采用两个指针，一前一后找到最大值存起来，如果当前的权值小于h，那么没必要计算，直接跳过，最后返回最大值。

三、代码

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