Leetcode Trapping Rain Water

題目：

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!

解題：

設置兩個指針從兩邊往中間走，每次把min(A[left], A[right])作爲下界牆，以left, right爲兩邊計算當前層的總面積，迭代下一層。

最後將算起來的總面積減去黑塊的面積。

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;
};