LeetCode: Permutations 解题报告

Permutations

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

SOLUTION 1:

经典的递归回溯题目，一次ACCEPT. 请也参考上一个题目LeetCode: Combinations 解题报告.

 1 public class Solution {
 2     public List<List<Integer>> permute(int[] num) {
 3         List<List<Integer>> ret = new ArrayList<List<Integer>>();
 4         if (num == null || num.length == 0) {
 5             return ret;
 6         }
 7
 8         dfs(num, new ArrayList<Integer>(), ret);
 9         return ret;
10     }
11
12     public void dfs(int[] num, List<Integer> path, List<List<Integer>> ret) {
13         int len = num.length;
14         if (path.size() == len) {
15             ret.add(new ArrayList<Integer>(path));
16             return;
17         }
18
19         for (int i = 0; i < len; i++) {
20             if (path.contains(num[i])) {
21                 continue;
22             }
23
24             path.add(num[i]);
25             dfs(num, path, ret);
26             path.remove(path.size() - 1);
27         }
28     }
29 }

SOLUTION 2:

可能有的同学觉得为什么path.contains不用hashmap来代替哩？所以主页君写了一个带hashmap的版本。结论是，在这个set规模小的时候，hashmap的性能还不

如arraylist。

原因可能在于，hashmap申请的不是一个连续的空间，而arraylist比较小的话，直接在连续内存中操作，速度会比较快。

以下是此程序的运行结果，hashmap的版本速度要慢一倍：

Test size:9
Computing time with HASHMAP: 629.0 millisec.
Test size:9
Computing time with list: 310.0 millisec.

 1 package Algorithms.permutation;
 2
 3 import java.util.ArrayList;
 4 import java.util.HashSet;
 5 import java.util.LinkedHashMap;
 6
 7 public class Permutation {
 8     public static void main(String[] strs) {
 9         int[] num = {1, 2, 3, 4, 5, 6, 7, 8, 9};
10         System.out.printf("Test size:%d \n", num.length);
11
12         Stopwatch timer1 = new Stopwatch();
13
14         permute(num);
15         System.out
16                 .println("Computing time with HASHMAP: "
17                         + timer1.elapsedTime() + " millisec.");
18
19         System.out.printf("Test size:%d \n", num.length);
20
21         Stopwatch timer2 = new Stopwatch();
22
23         permute2(num);
24         System.out
25                 .println("Computing time with list: "
26                         + timer2.elapsedTime() + " millisec.");
27     }
28
29     public static ArrayList<ArrayList<Integer>> permute(int[] num) {
30         ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
31         if (num == null) {
32             return ret;
33         }
34
35         permuteHelp(num, ret, new LinkedHashMap<Integer, Integer>());
36         return ret;
37     }
38
39     public static void permuteHelp(int[] num, ArrayList<ArrayList<Integer>> ret, LinkedHashMap<Integer, Integer> set) {
40         if (set.size() == num.length) {
41
42             ArrayList<Integer> list = new ArrayList<Integer>();
43             for (Integer i: set.keySet()){
44                 list.add(i);
45             }
46             ret.add(list);
47             return;
48         }
49
50         int len = num.length;
51         for (int i = 0; i < len; i++) {
52             if (set.containsKey(num[i])) {
53                 continue;
54             }
55
56             //path.add(num[i]);
57             set.put(num[i], 0);
58             permuteHelp(num, ret, set);
59             //path.remove(path.size() - 1);
60             set.remove(num[i]);
61         }
62     }
63
64     public static ArrayList<ArrayList<Integer>> permute2(int[] num) {
65         ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
66         if (num == null) {
67             return ret;
68         }
69
70         ArrayList<Integer> path = new ArrayList<Integer>();
71         permuteHelp2(num, path, ret);
72         return ret;
73     }
74
75     public static void permuteHelp2(int[] num, ArrayList<Integer> path, ArrayList<ArrayList<Integer>> ret) {
76         if (path.size() == num.length) {
77             ret.add(new ArrayList<Integer>(path));
78             return;
79         }
80
81         int len = num.length;
82         for (int i = 0; i < len; i++) {
83             if (path.contains(num[i])) {
84                 continue;
85             }
86
87             path.add(num[i]);
88             permuteHelp2(num, path, ret);
89             path.remove(path.size() - 1);
90         }
91     }
92 }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dfs/Permute.java