PermutationsUnique，求全排列，去重

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問題描述：給定一個數組，數組裡面有重複元素，求全排列。

演算法分析：和上一道題一樣，只不過要去重。

 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]交換，如果start-end之間有nums[i]==nums[end],那說明它以前交換過，就不用重複了。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     //這種方法和Permutation一樣，因為用set了，所以就已經去重了。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 }

 

PermutationsUnique，求全排列，去重