leetcode 11 Container With Most Water

题意：

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 and n is at least 2.

思路：

暴力计算所有的ai,aj组合，计算过程利用如下性质：记录当前取得max_area时的max_start和max_end，固定ai，从右遍历aj时，aj必须大于max_end才有可能超过当前max_area。外层循环从左开始遍历ai，只有ai大于max_start才有可能超过max_area。

代码：

class Solution {
public:
    int maxArea(vector<int>& height) {
        int max_area = 0;
        int max_end = 0;
        int max_start = 0;
        for(int i = 0;i < height.size();i++)
        {
            if(height[i] > max_start)
            {
                for(int j = height.size() - 1;j > i;j--)
                {
                    if(height[j] >= max_end)
                    {
                        int temp_area = (j - i) * min(height[i],height[j]);
                        if(temp_area > max_area)
                        {
                            max_start = height[i];
                            max_end = height[j];
                            max_area = temp_area;
                        }
                    }
                }
            }
        }
        return max_area;
    }
};

这个解法比较low，时间复杂度应该还是o(n²)，leetcode上o(n)的解法可参见：

https://leetcode.com/problems/container-with-most-water/solution/