LeetCode - Permutations II

Permutations II

2013.12.15 05:30

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].

Solution:

　　This time the array may contain duplicates, so I implemented the next_permutation() myself, and called it until all the permutations are traversed.

　　You can see the two problems here and find their connection with this problem: Next Permutation and Permutations.

　　Time complexity is O(n * n! / (a! * b! * c! * ...)), where a, b, c, ... are the number of groups of duplictate elements. Space complexity is O(n).

Accepted code:

 1 // 1CE, 2MLE, 1AC, not so easy

 2 #include <algorithm>

 3 using namespace std;

 4 

 5 class Solution {

 6 public:

 7     vector<vector<int> > permuteUnique(vector<int> &num) {

 8         // IMPORTANT: Please reset any member data you declared, as

 9         // the same Solution instance will be reused for each test case.

10         int i, j;

11         int n;

12         

13         n = result.size();

14         for(i = 0; i < n; ++i){

15             result[i].clear();

16         }

17         result.clear();

18         

19         sort(num.begin(), num.end());

20         n = num.size();

21         if(n <= 0){

22             return result;

23         }

24         while(true){

25             result.push_back(num);

26             for(i = n - 1; i > 0; --i){

27                 if(num[i - 1] < num[i]){

28                     break;

29                 }

30             }

31             if(i <= 0){

32                 break;

33             }

34             --i;

35             for(j = n - 1; j > i; --j){

36                 if(num[j] > num[i]){

37                     break;

38                 }

39             }

40             myswap(num[j], num[i]);

41             // 2MLE here, reverse(i + 1, n - 1), not reverse(i, n - 1);

42             reverse(num, i + 1, n - 1);

43         }

44         

45         return result;

46     }

47 private:

48     vector<vector<int>> result;

49     void myswap(int &x, int &y) {

50         int tmp;

51         

52         tmp = x;

53         x = y;

54         y = tmp;

55     }

56     

57     void reverse(vector<int> &num, int left, int right) {

58         int i;

59         

60         if(left > right){

61             reverse(num, right, left);

62             return;

63         }

64         

65         i = left;

66         // 1CE here, how did you submit before even finishing it!!!

67         while(i < left + right - i){

68             myswap(num[i], num[left + right - i]);

69             ++i;

70         }

71     }

72 };