leetcode #46 in cpp

Given a collection of distinct 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]

.
Solution: 

recurrent structure: 

permutation of [1,2,3] = [1 , permutation(2,3)]  and [2, permutation[1,3]] and [3, permutation(1,2)]; 

Code: 

class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> res;
vector<int> members;
find_permutation(members, nums, &res);
return res;
}

void find_permutation(vector<int> members, vector<int> nums, vector<vector<int>> *res){
if(nums.size() == 1){
members.push_back(nums[0]);
res->push_back(members);
return;
}
for(int i = 0; i < nums.size(); i ++){
int temp = nums[i];
members.push_back(temp);//separate one number out of the array
nums.erase(nums.begin() + i);
find_permutation(members, nums, res);//find the permutation of the rest of numbers
nums.insert(nums.begin() + i, temp);
members.pop_back();
}
};
};

If we modify the code a little bit, we could abandon the use of 'members' and use less memory by using an index indicating where the permutation begins.

Code : 

class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> res;
find_permutation(0, nums, &res);
return res;
}

void find_permutation(int start, vector<int> nums, vector<vector<int>> *res){
if(start == nums.size() - 1){
res->push_back(nums);
return;
}
for(int i = start; i < nums.size(); i ++){
swap(nums[i], nums[start]);
find_permutation(start + 1, nums, res);
swap(nums[i], nums[start]);
}
};

};