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], return6.

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!

解题思路：

类似于前述最大矩形解法，前为切头，此为填坑，异曲同工；

同理，可改为stack，复杂度为O(n): 将计算过的部分弹栈，认为其高度等于当前新水坝高度。

复杂度为O(n^2);

class Solution {public:    int minn( int a, int b){        if( a < b){            return a;        }        return b;    }    int trap(int A[], int n) {        //记录左侧水坝最大高度        //每次加入新水坝，判断是否满足蓄水条件（水坝高度是否大于前一个水坝）        //更新储水量，并填坑。                 //更新储水量：直至找到左侧最大高度的水坝或和当前水坝一样高的水坝        //填坑：为防止重复计算，将计算过的地方填平。        int sum = 0;        int maxLeft = A[0];        for(int i= 1 ; i< n ; i++ ){            //满足蓄水条件            if( A[i] > A[i-1] ){                int cur = minn(maxLeft,A[i]);                int cnt = 0;                int k = i-1;                while( A[k] < cur ){                    cnt -= A[k];                    A[k] = cur;                    k--;                }                   cnt += cur * (i - k -1);                sum += cnt;            }            //更新maxLeft；            if( A[i] > maxLeft){                maxLeft = A[i];            }        }        return sum;    }};