Given a collection of numbers, return all possible permutations.
For example,[1,2,3] have the following permutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
Analyse:
(1) Swap the 1st element with all the elements, including itself.
(2) Then the 1st element is fixed, go to the next element.
(3) Until the last element is fixed. Output.
Runtime: 16ms
1 class Solution { 2 public: 3 vector<vector<int>> permute(vector<int>& nums) { 4 vector<vector<int> > result; 5 if(nums.empty()) return result; 6 7 helper(result, nums, 0, nums.size() - 1); 8 return result; 9 } 10 11 void helper(vector<vector<int> >& result, vector<int> nums, int depth, int n){ 12 if(depth == n){ 13 result.push_back(nums); 14 return; 15 } 16 for(int i = depth; i < nums.size(); i++){ 17 swap(nums[depth], nums[i]); 18 helper(result, nums, depth + 1, n); 19 swap(nums[depth], nums[i]); 20 } 21 } 22 };
原文：http://www.cnblogs.com/amazingzoe/p/4873385.html