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Well, have you solved the nextPermutation problem? If so and you have handled the cases of duplicates at that problem, your code can be used in this problem. The idea is fairly simple:
nums in ascending order, add it to res;nums using nextPermutation(), and add it to res;nums returns to the sorted condition in 1.The code is as follows. For more about the idea of nextPermutation(), please visit this solution.
1 bool nextPermutation(vector<int>& nums) { 2 int k = -1; 3 for (int i = nums.size() - 2; i >= 0; i--) { 4 if (nums[i] < nums[i + 1]) { 5 k = i; 6 break; 7 } 8 } 9 if (k == -1) { 10 sort(nums.begin(), nums.end()); 11 return false; 12 } 13 int l = -1; 14 for (int i = nums.size() - 1; i > k; i--) { 15 if (nums[i] > nums[k]) { 16 l = i; 17 break; 18 } 19 } 20 swap(nums[k], nums[l]); 21 reverse(nums.begin() + k + 1, nums.end()); 22 return true; 23 } 24 vector<vector<int>> permuteUnique(vector<int>& nums) { 25 vector<vector<int> > res; 26 sort(nums.begin(), nums.end()); 27 res.push_back(nums); 28 while (nextPermutation(nums)) 29 res.push_back(nums); 30 return res; 31 }
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4547994.html
