[*leetcode 42] Trapping Rain Water

浏览数：27 / 时间：2015年06月11日

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

技术分享

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

[Solution]

由两边向中间搜索，

如下面的分析图，会比较直观：

技术分享

这里计算面积不用一般几何书的方法，这里是两边往中间遍历，记录当前第二高点secHight，然后利用这个第二高点减去当前历经的柱子，剩下就装水容量了。

为什么是第二高点？因为两边比较，最高的点不用动，只移动第二高点。

参考：http://www.xuebuyuan.com/1586534.html

 1 int trap(vector<int>& height) 
 2     {
 3         int left = 0, right = height.size() - 1, sechight = 0;
 4         int area = 0;
 5         while (left < right)
 6         {
 7             if (height[left] < height[right])
 8             {
 9                 sechight = max(height[left], sechight);
10                 area += sechight-height[left];  
11                 left++;
12             } 
13             else 
14             {
15                 sechight = max(height[right], sechight);
16                 area += sechight - height[right];
17                 right--;
18             }
19         }
20         return area;
21     }