Trapping Rain Water

標籤：

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!

思路：

1.先把最左邊的0去掉；

2.挨個遍曆每個柱子。

3.去重的處理

代碼：

class Solution{public:    int trap(int A[], int n) {        if(A==NULL||n==0) return 0;        int i;        int res=0;        for(i=0;i<n;++i){            if(A[i]!=0) break;        }        int start=i+1;        map<int,bool> repeatNum;        while (start<n)        {            if(repeatNum.count(start)){++start;continue;}            int l;int r;            for(l=start-1;l>=0;--l){                if(A[l]>A[start]){                    break;                }            }            for (r=start+1;r<n;++r)            {                if(A[r]==A[start]) repeatNum[r]=true;                if(A[r]>A[start]){                    break;                }            }            if(l==-1||r==n){                ++start;continue;            }            int temp=(min(A[l],A[r])-A[start])*(r-l-1);            res+=temp;            ++start;        }                return res;    }};

 

 

Trapping Rain Water