Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

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

和permutations 基本一致， 只是需要先把array 排序，然后保证在同一次循环中选取的数字不同即可。

 1 public class Solution {

 2     ArrayList<ArrayList<Integer>> result = null;

 3     int len = 0;

 4     public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {

 5         // Start typing your Java solution below

 6         // DO NOT write main() function

 7         Arrays.sort(num);

 8         result = new ArrayList<ArrayList<Integer>>();

 9         if(num == null || num.length == 0) return result;

10         len = num.length;

11         boolean[] item = new boolean[len];

12         get(item, num, new ArrayList<Integer>());

13         return result;

14     }

15     public void get(boolean[] item, int[] num, ArrayList<Integer> row){

16         if(row.size() == len){

17             result.add(new ArrayList<Integer>(row));

18             return;

19         }

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

21             if(!item[i]){

22                 item[i] = true;

23                 row.add(num[i]);

24                 get(item, num, row);

25                 item[i] = false;

26                 row.remove(row.size() - 1);

27                 while(i + 1 < num.length && num[i + 1] == num[i])  

28                     i ++;

29             }

30         }

31     }

32 }

第三遍： 不使用额外的boolean 数组：

 1 public class Solution {

 2     public List<List<Integer>> permuteUnique(int[] num) {

 3         ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();

 4         Arrays.sort(num);

 5         getPermutation(num, num.length, new ArrayList<Integer>(), result);

 6         return result;

 7     }

 8     public void getPermutation(int[] num, int len, ArrayList<Integer> row, ArrayList<List<Integer>> result){

 9         if(num == null || len == 0){

10             result.add(row);

11             return;

12         }

13         int pre = Integer.MAX_VALUE;

14         for(int i = 0; i < num.length; i ++){

15             if(num[i] != Integer.MAX_VALUE && num[i] != pre){

16                 pre = num[i];

17                 num[i] = Integer.MAX_VALUE;

18                 ArrayList<Integer> nrow = new ArrayList<Integer>(row);

19                 nrow.add(pre);

20                 getPermutation(num, len - 1, nrow, result);

21                 num[i] = pre;

22             }

23         }

24     }

25 }

 

第四遍：

 1 public class Solution {

 2     ArrayList<List<Integer>> result;

 3     public List<List<Integer>> permuteUnique(int[] num) {

 4         result = new ArrayList<List<Integer>>();

 5         if(num == null || num.length == 0) return result;

 6         Arrays.sort(num);

 7         findPermutation(num, new ArrayList<Integer>());

 8         return result;

 9     }

10     

11     private void findPermutation(int[] num, ArrayList<Integer> row){

12         if(row.size() == num.length){

13             result.add(row);

14         }

15         int pre = Integer.MIN_VALUE;

16         for(int i = 0; i < num.length; i ++){

17             if(num[i] != Integer.MIN_VALUE && num[i] != pre){

18                 pre = num[i];

19                 row.add(num[i]);

20                 num[i] = Integer.MIN_VALUE;

21                 findPermutation(num, new ArrayList<Integer>(row));

22                 num[i] = pre;

23                 row.remove(row.size() - 1);

24             }

25         }

26     }

27 }