LeetCode习题笔记——水箱问题

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.

装水问题其实就是找最短板的问题，解答如下：

class Solution {
public:
    int maxArea(vector<int> &height) {
    int start = 0;  
    int end = height.size()-1;  
    int maxV = 0;  
    while (start < end)  {  
        int contain = min(height[end], height[start]) * (end-start);  
        maxV = max(maxV, contain);  
        if (height[start] <= height[end])  {  
          start++;  
        }  
        else  {  
          end--;  
        }  
      }  
      return maxV;  
    }
};

因为水箱的容积是由短的一边乘上线段之间的距离得到的，所以当得出哪边短之后，就让这边的线段移动一次，即换一个边进行判断，之后便可得到最大的容积。