Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]
和 subset 差不多的解决方法,都是使用回溯法。
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> res;
vector<int> path;
dfs(nums,path,res);
return res;
}
void dfs(vector<int>& nums,vector<int>& path,vector<vector<int>>& res)
{
if(path.size()==nums.size()){
res.push_back(path);
return;
}
for(auto i:nums){
auto pos=find(path.begin(),path.end(),i);
if(pos==path.end()){
path.push_back(i);
dfs(nums,path,res);
path.pop_back();
}
}
}
};