LeetCode —— Trapping Rain Water

链接：http://leetcode.com/onlinejudge#question_42

原题：

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!

思路：

我是先将数组排序，从大到小。用左右两个left，right下标，从最大值开始，先两边扩散，

如果落在[left，right]之间的，就pass。

这样的复杂度是O（nlogn）的，主要是排序上面。

不过discussion里面有一个强大的O（n）方法。

代码：

bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
    return a.first > b.first;
}

class Solution {
public:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (n <= 2)
            return 0;
        
        vector<pair<int, int> > vec;
        for (int i=0; i<n; i++)
            vec.push_back(pair<int, int>(A[i], i));
        
        sort(vec.begin(), vec.end(), cmp);
        
        int total = 0;
        int left, right;
        left = right = vec[0].second;
        int count = 1;
        while (count < n) {
            int cur = vec[count].second;
            if (cur > left && cur < right) {
                count++;
                continue;
            }
            
            if (cur < left) {
                for (int i=cur+1; i<left; i++) {
                    total += (vec[count].first - A[i]);
                }
                left = cur;
            }
            
            if (cur > right) {
                for (int i=right+1; i<cur; i++) {
                    total += (vec[count].first - A[i]);
                }
                right = cur;
            }
            
            count++;
        }
        
        return total;
    }
};