[Leetcode] permutation sequence

Permutation sequence

The Set[1,2,3,…,n]Contains a totalN! Unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie,N= 3 ):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

 

GivenNAndK, ReturnKTh permutation sequence.

Note:GivenNWill be between 1 and 9 aggressive.

Algorithm ideas:

Thought 1:

The most intuitive idea is to find and count DFS one by one. Timeout after submission.

The Code is as follows:

 1 public class Solution { 2     boolean success = false; 3     StringBuilder str = new StringBuilder(); 4     int count = 0; 5     public String getPermutation(int n, int k) { 6         int count = 1; 7         for(int i = 1; i <= n; count*=i++); 8         if(k <= 0 || k > count ) return null; 9         boolean[] hash = new boolean[10];10         for(int i = 1; i <= n;hash[i++] = true);11         StringBuilder sb = new StringBuilder();12         dfs(sb,hash,k,n);13         return str.toString();14     }15     private void dfs(StringBuilder sb,boolean[] set,int k,int n){16         if(sb.length() == n){17             count++;18             if(count == k) {19                 success = true;20                 str = sb;21             }22             return;23         }24         for(int i = 1; i <= n ; i++){25             if(!set[i]) continue;26             set[i] = false;27             sb.append(i);28             dfs(sb, set, k, n);29             if(success) return;30             set[i] = true;31             sb.deleteCharAt(sb.length() - 1);32         }33     }34     35 }

View code

 

Idea 2:

Skip the useless ones in the middle and directly find the K.

Search rules:

Take n = 4, K = 20 as an example. There are 3 strings starting with 1! = 6, likewise, there are 6 at the beginning of 2 and 3, so 20th must start with 4.

Next, you only need to request 2nd (20-18) numbers starting with 4.

Recursive processing.

 1 public class Solution { 2     public String getPermutation(int n, int k) { 3         int count = 1; 4         for(int i = 1; i <= n; count*=i++); 5         if(k <= 0 || k > count ) return null; 6         List<Integer> list = new ArrayList<Integer>(); 7         for(int i = 1; i <= n; list.add(i++)); 8         StringBuilder sb = new StringBuilder(); 9         helper(n, k, list, sb);10         return sb.toString();11     }12     13     public void helper(int n,int k ,List<Integer> list,StringBuilder sb){14         if(n == 1) {15             sb.append(list.get(0));16             return;17         }18         int count = 1;19         for(int i = 1; i <= n - 1;count *= i++);20         int index = 0;21         while(k > count){22             index++;23             k -= count;24         }25         sb.append(list.get(index));26         list.remove(index);27         helper(n - 1,k,list,sb);28     }    29 }

 

 

Thought 3:

Iterative Processing

Transfer Gate