[leetcode]Permutation Sequence

Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

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

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

 

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

算法思路：

思路1：

最直观的思路，dfs逐个求，并计数。提交之后超时。

代码如下：

 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

 

思路2：

跳过中间那些无用的，直接求第k个。

寻找规律：

以n = 4,k = 20为例。以1为头的字符串一共有3!= 6个，同理以2、3开头的也有6个，因此第20个必定以4开头。

接下来只需要求以4开头的第2（20 - 18）个即可。

再递归处理。

 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 }

 

 

思路3：

迭代处理

传递门

posted on 2014-07-23 20:32  喵星人与汪星人  阅读(329)  评论(0)  编辑  收藏  举报