Leetcode Trapping Rain Water

题目：

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!

解题：

设置两个指针从两边往中间走，每次把min(A[left], A[right])作为下界墙，以left, right为两边计算当前层的总面积，迭代下一层。

最后将算起来的总面积减去黑块的面积。

class Solution {public:    int trap(int A[], int n) {        if(!n) return 0;        area = tot = 0;        int le = 0, ri = n - 1;        for(;le < n && !A[le]; le ++);        for(;ri >= le && !A[ri]; ri --);        for(int i = le; i <= ri; i ++)            tot += A[i];        gao(A, 0, n - 1, 0);        return area - tot;    }    void gao(int A[], int le, int ri, int pre) {        while(A[le] <= pre && le < ri) le ++;        while(A[ri] <= pre && le < ri) ri --;        if(le == ri) {            tot -= A[le] - pre;            return ;        }        area += (min(A[le], A[ri]) - pre) * (ri - le + 1);        gao(A, le, ri, min(A[le], A[ri]));        return ;    }private:    int area, tot;};