leetcode : 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. 

 思路同在直方图中找最大矩形。维护一个递减的栈s，如果A[i]大于s.top()，一直出栈直A[i]<s.top()或者栈空。注意的是算面积时，每次出栈相当于把下面的坑填平。

AC代码：

class Solution {
public:
    int trap(int A[], int n) {
        stack<int> s;
        int sum = 0;
        int base = A[0];          //记录栈底的值
        for(int i = 0; i < n; ++i){
            if(s.empty() || A[i] < A[s.top()]){
                s.push(i);
            }else{
                int maxVal = A[i];
                if(A[i] > base){   //确定边界值
                    maxVal = base;
                    base = A[i];
                }
                while(!s.empty() && A[i] > A[s.top()]){
                    int tempVal = A[s.top()];
                    int tempIndex = s.top();
                    s.pop();
                    if(tempVal == maxVal){  //如果栈要空了，那么最后一个元素当成边界使用
                        break;
                    }
                    sum += (maxVal - tempVal) * (tempIndex - s.top());
                }
                s.push(i);
            }
        }
        return sum;
    }
};