[leetcode]Trapping Rain Water

標籤：leetcode   演算法   

問題描述：

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.

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!

基本思路：

本題通過分析可以發現，儲存水的槽的兩邊都是槽高的極大值。(這裡對於最左邊的曹，只要大於等於它右面的即可。最右邊的同理)

但是單純的考量 每兩個極大值的槽作為水槽的兩邊得到的結果是不對的。舉個例子：

A = {9，2, 9, 2,2,1,8}

這裡如果只考慮每兩個極大值之間存放的水，結果是7+0+1 = 8. 而實際上最大的結果應該是7+6+6+7 = 26.

所以，以沒有計算過的最左的槽為左槽，找到符合以下條件之一的槽為右槽：

1. 該槽的槽高>= 最左的槽高
2. 該槽是最左槽右側槽中槽高最大的。
   

(PS:此題我寫了好久。。。  總是有地方出錯。  後來發現是局部極大值和局部極大值的位置搞錯了-.-|| )

代碼：

    int trap(int A[], int n) { //C++        if(n < 3)            return 0;       vector<int> locmaxs;                int sum = 0;        int i = 0;        while(i < n)        {                if(i == 0 && A[0] >= A[1])                {                    locmaxs.push_back(0);                    i++;                                    }                else if(i == n-1 && A[n-1] >= A[n-2])                {                    locmaxs.push_back(n-1);                    i++;                                                                                   }                else if(A[i] >= A[i-1] && A[i] >= A[i+1])                {                    locmaxs.push_back(i);                    i++;                }                else i++;        }                        for(i = 0; i < locmaxs.size()-1; )        {            int pos = i+1 ,max = locmaxs[i+1];            for( int j = i+1;  j < locmaxs.size(); j++)            {                if(A[locmaxs[j]] >= A[locmaxs[i]])                {                    pos = j;                    break;                }                if(A[locmaxs[j]] > A[max])                {                    max = locmaxs[j];                    pos = j;                }            }                        int draw = ((A[locmaxs[i]] > A[locmaxs[pos]]) ? A[locmaxs[pos]]: A[locmaxs[i]]);            for(int j = locmaxs[i]+1; j < locmaxs[pos]; j++)                if(draw - A[j] >0)                    sum += draw - A[j];            i = pos;        }        return sum;    }

[leetcode]Trapping Rain Water