LeetCode Container With Most Water

class Solution{

    public:

        int maxArea(vector<int>& height) {  

        int len = height.size(), low = 0, high = len -1 ;  

        int maxArea = 0;  

        while (low < high) {  

            maxArea = max(maxArea, (high - low) * min(height[low], height[high]));  

            if (height[low] < height[high]) {  

                low++;  

            } else {  

                high--;  

            }  

        }  

        return maxArea;  

    }

};

应该算是贪婪算法吧，直接从disscus里抄了，自己只能想出一个nlogn的，对贪婪没什么感觉，想不到

第二轮：

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.

以前leetcode还没有标flag，现在可以看到这题标了two pointer就是2sum这种类似，首先最大的可能是由两端的作为容器壁，然后慢慢往内收缩尝试，收缩过程中放弃比较低的容器壁。反正这次又没想出来还看错题目，智商拙计啊：

 1 // 9:44

 2 class Solution {

 3 public:

 4     int maxArea(vector<int>& height) {

 5         int len = height.size();

 6         int lo = 0, hi = len - 1;

 7         int maxarea = 0;

 8         while (lo < hi) {

 9             maxarea = max(maxarea, min(height[lo], height[hi]) * (hi - lo));

10             if (height[lo] < height[hi]) {

11                 lo++;

12             } else {

13                 hi--;

14             }

15         }

16         

17         return maxarea;

18     }

19 };