Trapping Rain Water

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!

 

 第一遍：
 1 public class Solution {
 2     public int trap(int[] A) {
 3         if(A == null || A.length < 3) return 0;
 4         int start = 0, end = A.length - 1;
 5         while(A[start] == 0){
 6             start ++;
 7         }
 8         while(A[end] == 0){
 9             end --;
10         }
11         int left = start;
12         int sum = 0;
13         for(int i = start + 1; i <= end; i ++){
14             if(A[i] >= A[left]){
15                 for(int j = left + 1; j < i; j ++){
16                     sum += A[left] - A[j];
17                 }
18                 left = i;
19             }
20         }
21         int right = end;
22         for(int i = end - 1; i >= start; i --){
23             if(A[i] > A[right]){
24                 for(int j = i + 1; j < right; j ++){
25                     sum += A[right] - A[j];
26                 }
27                 right = i;
28             }
29         }
30         return sum;
31     }
32 }

 

第三遍：

 1 public class Solution {
 2     public int trap(int[] A) {
 3         if(A == null || A.length < 3) return 0;
 4         int sum = 0;
 5         int start = 0, end = A.length - 1;
 6         while(start < A.length && A[start] == 0) start ++;
 7         while(end > -1 && A[end] == 0) end --;
 8         int per = start, cur = start + 1;
 9         for(; cur <= end; cur ++){
10             if(A[cur] >= A[per]){
11                 for(int i = per + 1; i < cur; i ++)
12                     sum += A[per] - A[i];
13                 per = cur;
14             }
15         }
16         per = end; cur = end - 1;
17         for(; cur >= start; cur --){
18             if(A[cur] > A[per]){
19                 for(int i = cur + 1; i < per; i ++)
20                     sum += A[per] - A[i];
21                 per = cur;
22             }
23         }
24         return sum;
25     }
26 }

 

posted on 2014-08-15 09:38  Step-BY-Step  阅读(126)  评论(0)  编辑  收藏  举报

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