问题描述：

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：

本题通过分析可以发现，存储水的槽的两边都是槽高的极大值。(这里对于最左边的曹，只要大于等于它右面的即可。最右边的同理)

但是单纯的考量 每两个极大值的槽作为水槽的两边得到的结果是不对的。举个例子：

A = {9，2, 9, 2,2,1,8}

这里如果只考虑每两个极大值之间存放的水，结果是7+0+1 = 8. 而实际上最大的结果应该是7+6+6+7 = 26.

所以，以没有计算过的最左的槽为左槽，找到符合以下条件之一的槽为右槽：

1. 该槽的槽高>= 最左的槽高
2. 该槽是最左槽右侧槽中槽高最大的。
   

(PS:此题我写了好久。。。  总是有地方出错。  后来发现是局部极大值和局部极大值的位置搞错了-.-|| )

代码：

    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;
    }