# A little Change With Permutation I can solve this problem

• Permutation II:

``````class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> ret;
recursion(nums,0,ret);
return ret;
}
void recursion(vector<int> &nums,int index,vector<vector<int>> &ret){
if(index == nums.size()-1){
ret.push_back(nums);
return;
}
for(int i=index; i<nums.size(); ++i){
if(dupsolution(nums,index,i))
continue;
swap(nums[index],nums[i]);
recursion(nums,index+1,ret);
swap(nums[i],nums[index]);
}
}
void swap(int &a,int &b){
int temp = a;
a = b;
b = temp;
}
bool dupsolution(vector<int> nums,int step,int end){
//if in range [index,j),there exists number equals num[j],the swap with j is a dupsolution,we swap them early.so we pass by
for(int i=step; i<end; ++i){
if(nums[i] == nums[end])
return true;
}
return false;
}
};
``````

Permuation I

``````class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> ret;
recursion(nums,0,ret);
return ret;
}
void recursion(vector<int> &nums,int index,vector<vector<int>> &ret){
if(index == nums.size()-1){
ret.push_back(nums);
return;
}
for(int i=index; i<nums.size(); ++i){
swap(nums[index],nums[i]);
recursion(nums,index+1,ret);
swap(nums[i],nums[index]);
}
}
void swap(int &a,int &b){
int temp = a;
a = b;
b = temp;
}
};``````

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