LeetCode --- 46. Permutations

LeetCode --- 46. Permutations

Link: Permutations

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]. 

This question requires that a group of numbers be given to generate all the permutation and combination.

1. Recursive Arrangement

This is a problem of arrangement. The first thing we think of is recursion. Fix each element one by one to the first position, and recursively arrange the remaining elements. When the number of elements fixed to the front is equal to the length of the array, recursion ends.

Time Complexity: O (n !) (Number of results)

Space complexity: O (n !)

 1 class Solution 2 { 3 public: 4     vector
  
    > permute(vector
   
     &num) 5     { 6         vector
    
      > vvi; 7         permute(num, 0, vvi); 8         return vvi; 9     }10 private:11     void permute(vector
     
       &num, int i, vector
      
        > &vvi)12 {13 if(i == num.size())14 {15 vvi.push_back(num);16 return;17 }18 for(int j = i; j < num.size(); ++ j)19 {20 swap(num[i], num[j]);21 permute(num, i + 1, vvi);22 swap(num[i], num[j]);23 }24 }25 };
      
     
    
   
  

2. next_permutation

In fact, you can also use the next_permutation () function to gradually generate the next arrangement. Because next_permutation () returns false in the last arrangement, you can first sort the array and then call next_permutation () until it returns false.

Time Complexity: O (n !) (Number of results)

Space complexity: O (n !)

 1 class Solution 2 { 3 public: 4     vector
  
    > permute(vector
   
     &num) 5     { 6         sort(num.begin(), num.end()); 7          8         vector
    
      > vvi({num}); 9         while(next_permutation(num.begin(), num.end()))10             vvi.push_back(num);11         12         return vvi;13     }14 };