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题目：
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!
代码：
class Solution { public: int trap(vector<int>& height) { // non valid input int len = height.size(); if (len<=2) return 0; // initial int left_max[height.size()]; int right_max[height.size()]; left_max[0] = 0; right_max[len-1] = 0; // get left_max and right_max for (int i = 1; i < len; ++i) { left_max[i] = std::max(left_max[i-1], height[i-1]); right_max[len-i-1] = std::max(right_max[len-i], height[len-i]); } // calculate the sum int sum = 0; for (int i = 0; i < len; ++i) { int h = std::min(left_max[i], right_max[i]); if (h>height[i]) { sum += h-height[i]; } } return sum; } };
Tips:
1. 遍历，获得每个位置上左边最高的和右边最高的；选择左边和右边比较小的高度，减去该位置的高度，就是可需水量。
2. 注意一些极端case的处理
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4457356.html
