Trapping Rain Water leetcode java

时间：2014-07-30 09:47:13 阅读：272 评论：0 收藏：0 [点我收藏+]

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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.

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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!

题解：

参考：低调小一（http://blog.csdn.net/wzy_1988/article/details/17752809）的解题思路

“首先，碰到这样的题目不要慌张，挨个分析每个A[i]能trapped water的容量，然后将所有的A[i]的trapped water容量相加即可

其次，对于每个A[i]能trapped water的容量，取决于A[i]左右两边的高度（可延展）较小值与A[i]的差值，即volume[i] = [min(left[i], right[i]) - A[i]] * 1，这里的1是宽度，如果the width of each bar is 2,那就要乘以2了”

那么如何求A[i]的左右高度呢？要知道，能盛多少水主要看短板。那么对每个A[i]来说，要求一个最高的左短板，再求一个最高的右短板，这两个直接最短的板子减去A[i]原有的值就是能成多少水了。

所以需要两遍遍历，一个从左到右，找最高的左短板；一个从右到左，找最高的右短板。最后记录下盛水量的总值就是最终结果了。

代码如下：

1      public int trap(int[] A) {
2         if (A == null || A.length == 0)
3             return 0;
4
5         int i, max, total = 0;
6         int left[] = new int[A.length];
7         int right[] = new int[A.length];
8
9         // from left to right
10         left[0] = A[0];
11         max = A[0];
12         for (i = 1; i < A.length; i++) {
13             left[i] = Math.max(max, A[i]);
14             max = Math.max(max, A[i]);
15         }
16
17         // from right to left
18         right[A.length-1] = A[A.length-1];
19         max = A[A.length-1];
20         for (i = A.length-2; i >= 0; i--) {
21             right[i] = Math.max(max, A[i]);
22             max = Math.max(max, A[i]);
23         }
24
25         // trapped water (when i==0, it cannot trapped any water)
26         for (i = 1; i < A.length-1; i++) {
27             int bit = Math.min(left[i], right[i]) - A[i];
28             if (bit > 0)
29                 total += bit;
30         }
31
32         return total;
33     }

对照着代码再看原来的例子：

index: 0 1 2 3 4 5 6 7 8 9 10 11

A[index]: 0 1 0 2 1 0 1 3 2 1 2 1

left[index]: 0 1 1 2 2 2 2 3 3 3 3 3

right[index]: 3 3 3 3 3 3 3 3 2 2 2 1

min[i]: 0 1 1 2 2 2 2 3 2 2 2 1

bit[i]: - 0 1 0 1 2 1 0 0 1 0 0

那么根据上表可以算出最终结果是6。

Reference：http://blog.csdn.net/wzy_1988/article/details/17752809

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Trapping Rain Water leetcode java

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