LeetCode – Refresh – Trapping Rain Water

weixin_33989780

于 2015-03-24 17:06:00 发布

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原文链接：http://www.cnblogs.com/shuashuashua/p/4363374.html

本文介绍了一种解决雨水陷阱问题的有效算法。通过双向扫描数组，分别从左到右及从右到左找到最大高度，利用这些信息计算可以积累的雨水量。文章提供了详细的C++实现代码。

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Scanning from left to right and right to left.

Find the max height of left and right. The minimum one will be most possible height for current state that can bound the water. But one thing need to check:

whether current height is higher than the value.

 1 class Solution {
 2 public:
 3     int trap(int A[], int n) {
 4         if (n < 3) return 0;
 5         vector<int> left(n), right(n);
 6         int result = 0, rec = A[0];
 7         for (int i = 1; i < n; i++) {
 8             left[i] = max(rec, left[i-1]);
 9             rec = max(rec, A[i]);
10         }
11         rec = A[n-1];
12         for (int i = n-2; i >= 0; i--) {
13             right[i] = max(rec, right[i+1]);
14             rec = max(rec, A[i]);
15         }
16         for (int i = 0; i < n; i++) {
17             result += (min(left[i], right[i]) - A[i]) > 0 ? (min(left[i], right[i]) - A[i]) : 0;
18         }
19         return result;
20     }
21 };