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：

// 9:44
class Solution {
public:
    int maxArea(vector<int>& height) {
        int len = height.size();
        int lo = 0, hi = len - 1;
        int maxarea = 0;
        while (lo < hi) {
            maxarea = max(maxarea, min(height[lo], height[hi]) * (hi - lo));
            if (height[lo] < height[hi]) {
                lo++;
            } else {
                hi--;
            }
        }
        
        return maxarea;
    }
};