[Leetcode] 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].

 

注意不能用n！来判断了，可以先排一下序，当序列逆序时结束。

 1 class Solution {

 2 public:

 3     bool nextPermutation(vector<int> &num) {

 4         if (num.size() < 2) {

 5             return false;

 6         }

 7         int i, k;

 8         bool flag = true;

 9         for (i = num.size() - 2; i >= 0; --i) {

10             if (num[i] < num[i + 1]) {

11                 break;

12             }

13         }

14         if (i < 0) {

15             flag = false;

16         }

17         for (k = num.size() - 1; k > i; --k) {

18             if (num[k] > num[i]) {

19                 break;

20             }

21         }

22         swap(num[i], num[k]);

23         reverse(num.begin() + i + 1, num.end());

24         return flag;

25     }

26     

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

28         vector<vector<int> > res;

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

30         res.push_back(num);

31         while (nextPermutation(num)) {

32             res.push_back(num);

33         }

34         return res;

35     }

36 };