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].
1 class Solution { 2 public: 3 vector<vector<int> > permute(vector<int> &num) { 4 vector<vector<int> > result; 5 sort(num.begin(), num.end()); 6 vector<int> path; 7 dfs(num, path, result); 8 return result; 9 } 10 private: 11 void dfs(vector<int> &num, vector<int> &path, vector<vector<int>> &result) { 12 if (path.size() == num.size()) { 13 result.push_back(path); 14 return; 15 } 16 for (auto i : num) { 17 if (find(path.begin(), path.end(), i) == path.end()) { 18 path.push_back(i); 19 dfs(num, path, result); 20 path.pop_back(); 21 } 22 } 23 } 24 };
可以利用next permutation的方法，一个一个求，也可以用dfs的方法来求所有的排列。上面的代码是dfs版。
【Leetcode】Permutations,布布扣,bubuko.com
原文：http://www.cnblogs.com/dengeven/p/3608081.html
