LeetCode题解——Trapping Rain Water

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!

这道题和container with most water 有点类似，但是这道题更难一点；分析题目可看出，随意给定一组数字，可能形成很多个"坑"，那么是否可以每次计算出一个坑中的水，最后将其相加得到最终结果。

用递归是容易做到的，每次找出两个最高的bar，然后计算这两个bar之间的水量，再分别在其左边和右边递归的进行寻找。

class Solution {public:    int trap(vector<int>& height) {//用递归可以实现                if(height.size()<3) return 0;        int maxidxI=0,maxidxJ=height.size()-1;        int temp,water = 0;for(int i=1;i<height.size()-1;i++){if(height[i]>height[maxidxI])maxidxI = i;}for(int i =height.size()-1;i>=0;i--){    if(height[i]>height[maxidxJ] && i!=maxidxI)        maxidxJ = i;}if(maxidxI>maxidxJ){int t = maxidxI;maxidxI =maxidxJ;maxidxJ =t;}        temp =0;        for(int m = maxidxI+1 ; m<= maxidxJ-1; m++)            temp+= height[m];        water+=(height[maxidxI]>height[maxidxJ]?height[maxidxJ]:height[maxidxI])*(maxidxJ - maxidxI -1) - temp;        vector<int> l(height.begin(),height.begin()+maxidxI+1),r(height.begin()+maxidxJ,height.end());                return water+trap(l)+trap(r);    }};