[LintCode] 带重复元素的排列

递归实现：

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers.
 5      * @return: A list of unique permutations.
 6      */
 7     vectorint> > permuteUnique(vector<int> &nums) {
 8         // write your code here
 9         sort(nums.begin(), nums.end());
10         vectorint> > permutations;
11         if (nums.empty()) return permutations;
12         permutate(nums, 0, permutations);
13         return permutations;
14     }
15 private:
16     void permutate(vector<int> nums, int start, vectorint> >& permutations) {
17         if (start == nums.size()) {
18             permutations.push_back(nums);
19             return;
20         }
21         for (int i = start; i < (int)nums.size(); i++) {
22             if (i == start || nums[i] != nums[start]) {
23                 swap(nums[start], nums[i]);
24                 permutate(nums, start + 1, permutations);
25             }
26         }
27     }
28 };

非递归实现（基于nextPermutation）：

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers.
 5      * @return: A list of unique permutations.
 6      */
 7     vectorint> > permuteUnique(vector<int> &nums) {
 8         // write your code here
 9         vectorint> > permutations;
10         if (nums.empty()) return permutations;
11         vector<int> copy(nums.begin(), nums.end());
12         nextPermutation(nums);
13         permutations.push_back(nums);
14         while (nums != copy) {
15             nextPermutation(nums);
16             permutations.push_back(nums);
17         }
18         return permutations;
19     }
20 private:
21     void nextPermutation(vector<int>& nums) {
22         int k = -1, n = nums.size();
23         for (int i = n - 2; i >= 0; i--) {
24             if (nums[i] < nums[i + 1]) {
25                 k = i;
26                 break;
27             }
28         }
29         if (k == -1) {
30             reverse(nums.begin(), nums.end());
31             return;
32         }
33         int l;
34         for (int i = n - 1; i > k; i--) {
35             if (nums[i] > nums[k]) {
36                 l = i;
37                 break;
38             }
39         }
40         swap(nums[l], nums[k]);
41         reverse(nums.begin() + k + 1, nums.end());
42     }
43 };