42. 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.思路：    通过累积求总和；从左右两边开始遍历，标记遍历过程中的左右两边的最大值，maxL,maxR,遍历的过程中保持两侧的最大高度；左右位置进行比较，然后判断是进行左边或右边操作(操作高度小的)；若果是左边操作，则左侧值height[left]与maxL进行比较然后进行更新maxL的值或sum值；右侧同理.*/#include <stdio.h>#include <iostream>#include <string>#include <sstream>#include <vector>#include <algorithm>using namespace std;class Solution {public:    int trap(vector<int>& height) {        int left=0,right=height.size()-1;int sum=0,maxL=0,maxR=0;while(left<right){if(height[left]<=height[right]){if(height[left]>=maxL) maxL=height[left];else sum+=maxL-height[left];left++;}else{if(height[right]>=maxR) maxR=height[right];else sum+=maxR-height[right];right--;}}return sum;    }};int main(){    Solution mys;vector<int> candidates;//[0,1,0,2,1,0,1,3,2,1,2,1]candidates.push_back(0);candidates.push_back(1);candidates.push_back(0);candidates.push_back(2);candidates.push_back(1);candidates.push_back(0);candidates.push_back(1);candidates.push_back(3);candidates.push_back(2);candidates.push_back(1);candidates.push_back(2);candidates.push_back(1);int res=mys.trap(candidates);cout<<res<<endl;    return 0;}