算法描述：
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
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!
Example:
Input: [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6
解题思路：两个指针分别从两端开始扫，直到相遇。再用额外的两个变量标记两端最高位，当当前指针所指变量的值低于最高变量，则增加结果。
int trap(vector<int>& height) { if(height.size()<2) return 0; int left = 0; int right = height.size()-1; int maxLeft =0; int maxRight = 0; int res = 0; while(left < right){ if(height[left] < height[right]){ if(height[left] > maxLeft) maxLeft = height[left]; else res += maxLeft- height[left]; left++; }else{ if(height[right] > maxRight) maxRight = height[right]; else res += maxRight - height[right]; right--; } } return res; }
LeetCode-42-Trapping Rain Water
原文：https://www.cnblogs.com/nobodywang/p/10363159.html