LeetCode: Trapping Rain Water 解题报告

https://oj.leetcode.com/problems/trapping-rain-water/

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!

SOLUTION 1:

从左到右扫描，计算到从左边到curr的最高的bar,从右到左扫描，计算到从右边到curr的最高的bar。

再扫描一次，把这两者的低者作为{桶}的高度，如果这个桶高于A[i]的bar,那么A[i]这个bar上头可以存储height - A[i]这么多水。把这所有的水加起来即可。

 1 public class Solution {

 2     public int trap(int[] A) {

 3         if (A == null) {

 4             return 0;

 5         }

 6         

 7         int max = 0;

 8         

 9         int len = A.length;

10         int[] left = new int[len];

11         int[] right = new int[len];

12         

13         // count the highest bar from the left to the current.

14         for (int i = 0; i < len; i++) {

15             left[i] = i == 0 ? A[i]: Math.max(left[i - 1], A[i]);

16         }

17         

18         // count the highest bar from right to current.

19         for (int i = len - 1; i >= 0; i--) {

20             right[i] = i == len - 1 ? A[i]: Math.max(right[i + 1], A[i]);

21         }

22         

23         // count the largest water which can contain.

24         for (int i = 0; i < len; i++) {

25             int height = Math.min(right[i], left[i]);

26             if (height > A[i]) {

27                 max += height - A[i];

28             }

29         }

30         

31         return max;

32     }

33 }

View Code

2015.1.14 redo:

合并2个for 循环，简化计算。

 

 1 public class Solution {

 2     public int trap(int[] A) {

 3         // 2:37

 4         if (A == null) {

 5             return 0;

 6         }

 7         

 8         int len = A.length;

 9         int[] l = new int[len];

10         int[] r = new int[len];

11         

12         for (int i = 0; i < len; i++) {

13             if (i == 0) {

14                 l[i] = A[i];

15             } else {

16                 l[i] = Math.max(l[i - 1], A[i]);

17             }

18         }

19         

20         int water = 0;

21         for (int i = len - 1; i >= 0; i--) {

22             if (i == len - 1) {

23                 r[i] = A[i];

24             } else {

25                 // but: use Math, not max

26                 r[i] = Math.max(r[i + 1], A[i]);

27             }

28             

29             water += Math.min(l[i], r[i]) - A[i];

30         }

31         

32         return water;        

33     }

34 }

View Code

 

 

 

CODE:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/Trap.java