Permutation Sequence, Solution

 

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):
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.

 1     String getPermutation(int n, int k) {  
 2         Vector<Integer> num = new Vector<Integer>(n);
 3         StringBuffer out = new StringBuffer();
 4         int max = 1;
 5         for(int i = 0; i < n; i ++){
 6             num.add(i + 1);
 7             max *= i + 1;
 8         }
 9         int choosed = 0;
10         k -= 1;
11         for(int i = 0; i < n; i ++){
12             max /= (n - i);
13             choosed = k / max;
14             out.append(num.elementAt(choosed));
15             num.remove(choosed);
16             out.append(" ");
17             k = k % max;
18         }
19         
20         return out.toString();
21     }

 

posted on 2013-04-15 07:29  Step-BY-Step  阅读(211)  评论(0)  编辑  收藏  举报

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