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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.
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：
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 }
[*leetcode 42] Trapping Rain Water
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原文地址：http://www.cnblogs.com/ym65536/p/4428815.html
