46. Permutations
Medium
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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]
]
Accepted
459,980
Submissions
788,865
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
sort(nums.begin(),nums.end());//如果要找全排列一开始就要从小到大排序
int n=nums.size();
int a[n]={};
for(int i=0;i<n;i++){
a[i]=nums[i];
}
vector<vector<int>> rt;
do{
vector<int> temp;
for(int i=0;i<n;i++){
temp.push_back(a[i]);
//cout<<temp[i]<<' ';
}
//cout<<endl;
rt.push_back(temp);
}while(next_permutation(a,a+n));
return rt;
}
};
dfs回溯这个方法要自己好好理解层与层之间的回溯关系,不然会很混乱
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> rt;
vector<int> temp;
vector<int> visited(nums.size(),0);
dfs(rt,temp,visited,nums);
return rt;
}
private:
void dfs(vector<vector<int>>& rt,vector<int>& temp,vector<int>& visited,vector<int>& nums){
if(temp.size()==nums.size()){
rt.push_back(temp);
return;
}
for(int i=0;i<nums.size();i++){
if(!visited[i]){
temp.push_back(nums[i]);
visited[i]=1;
dfs(rt,temp,visited,nums);
temp.pop_back();
visited[i]=0;
}
}
}
};