[LeetCode] Permutations
Well, have you solved the nextPermutation problem? If so, your code can be used in this problem. The idea is fairly simple:
- sort
nums
in ascending order, add it tores
; - generate the next permutation of
nums
usingnextPermutation()
, and add it tores
; - repeat 2 until the next permutation of
nums
returns to the sorted condition in 1.
The code is as follows.
A final note, the following code can be applied to the problem of Permutations II without any modification since the cases of duplicates have already been handled in nextPermutation()
. If you want to learn more about 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>> permute(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 }
Of course, you are supposed to use the backtracking idea. A simple code is as follows.
1 void permutate(vector<int> nums, int start, vector<vector<int> >& res) { 2 if (start == nums.size()) { 3 res.push_back(nums); 4 return; 5 } 6 for (int i = start; i < nums.size(); i++) { 7 swap(nums[i], nums[start]); 8 permutate(nums, start + 1, res); 9 } 10 } 11 12 vector<vector<int> > permute(vector<int>& nums) { 13 vector<vector<int> > res; 14 permutate(nums, 0, res); 15 return res; 16 }