PermutationsUnique，求全排列，去重

Posted 2020-07-22 32ddd

问题描述：给定一个数组，数组里面有重复元素，求全排列。

算法分析：和上一道题一样，只不过要去重。

 3 import java.util.ArrayList;
 4 import java.util.HashSet;
 5 import java.util.List;
 6 import java.util.Set;
 7 
 8 public class PermutationsUnique {
 9     public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
10         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
11         permuteUnique(num, 0, result);
12         return result;
13     }
14      
15     private void permuteUnique(int[] num, int start, ArrayList<ArrayList<Integer>> result) {
16      
17         if (start >= num.length ) {
18             ArrayList<Integer> item = convertArrayToList(num);
19             result.add(item);
20         }
21      
22         for (int j = start; j < num.length; j++) {
23             if (containsDuplicate(num, start, j)) {
24                 swap(num, start, j);
25                 permuteUnique(num, start + 1, result);
26                 swap(num, start, j);
27             }
28         }
29     }
30      
31     private ArrayList<Integer> convertArrayToList(int[] num) {
32         ArrayList<Integer> item = new ArrayList<Integer>();
33         for (int h = 0; h < num.length; h++) {
34             item.add(num[h]);
35         }
36         return item;
37     }
38     //nums[start]和nums[end]交换，如果start-end之间有nums[i]==nums[end],那说明它以前交换过，就不用重复了。
39     private boolean containsDuplicate(int[] arr, int start, int end) {
40         for (int i = start; i < end; i++) {
41             if (arr[i] == arr[end]) {
42                 return false;
43             }
44         }
45         return true;
46     }
47      
48     private void swap(int[] a, int i, int j) {
49         int temp = a[i];
50         a[i] = a[j];
51         a[j] = temp;
52     }
53     
54     
55     
56     
57     //这种方法和Permutation一样，因为用set了，所以就已经去重了。
58     public static List<List<Integer>> permuteUnique2(int[] num) {
59         List<List<Integer>> returnList = new ArrayList<>();
60         returnList.add(new ArrayList<Integer>());
61      
62         for (int i = 0; i < num.length; i++) {
63             Set<ArrayList<Integer>> currentSet = new HashSet<>();
64             for (List<Integer> l : returnList) {
65                 for (int j = 0; j < l.size() + 1; j++) {
66                     l.add(j, num[i]);
67                     ArrayList<Integer> T = new ArrayList<Integer>(l);
68                     l.remove(j);
69                     currentSet.add(T);
70                 }
71             }
72             returnList = new ArrayList<>(currentSet);
73         }
74      
75         return returnList;
76     }
77 
78     public static void main(String[] args)
79     {
80         Permutations pt = new Permutations();
81         int[] num = {1,2,1,3};
82         System.out.println(pt.permute(num).size());
83         System.out.println(pt.permute(num));
84     }
85 }