[LintCode] 带重复元素的排列

递归实现：

 class Solution {
 public:
     /**
      * @param nums: A list of integers.
      * @return: A list of unique permutations.
      */
     vector<vector<int> > permuteUnique(vector<int> &nums) {
         // write your code here
         sort(nums.begin(), nums.end());
         vector<vector<int> > permutations;
         if (nums.empty()) return permutations;
         permutate(nums, , permutations);
         return permutations;
     }
 private:
     void permutate(vector<int> nums, int start, vector<vector<int> >& permutations) {
         if (start == nums.size()) {
             permutations.push_back(nums);
             return;
         }
         for (int i = start; i < (int)nums.size(); i++) {
             if (i == start || nums[i] != nums[start]) {
                 swap(nums[start], nums[i]);
                 permutate(nums, start + , permutations);
             }
         }
     }
 };

非递归实现（基于nextPermutation）：

 class Solution {
 public:
     /**
      * @param nums: A list of integers.
      * @return: A list of unique permutations.
      */
     vector<vector<int> > permuteUnique(vector<int> &nums) {
         // write your code here
         vector<vector<int> > permutations;
         if (nums.empty()) return permutations;
         vector<int> copy(nums.begin(), nums.end());
         nextPermutation(nums);
         permutations.push_back(nums);
         while (nums != copy) {
             nextPermutation(nums);
             permutations.push_back(nums);
         }
         return permutations;
     }
 private:
     void nextPermutation(vector<int>& nums) {
         int k = -, n = nums.size();
         for (int i = n - ; i >= ; i--) {
             if (nums[i] < nums[i + ]) {
                 k = i;
                 break;
             }
         }
         if (k == -) {
             reverse(nums.begin(), nums.end());
             return;
         }
         int l;
         for (int i = n - ; i > k; i--) {
             if (nums[i] > nums[k]) {
                 l = i;
                 break;
             }
         }
         swap(nums[l], nums[k]);
         reverse(nums.begin() + k + , nums.end());
     }
 };