LeetCode: Permutation Sequence 解题报告

Permutation Sequence

https://oj.leetcode.com/problems/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):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the k^th permutation sequence.

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

解答：

1. 以某一数字开头的排列有(n-1)! 个。
例如： 123， 132，以1开头的是 2！个
2. 所以第一位数字就可以用（k-1） / (n-1)! 来确定 .这里K-1的原因是，序列号我们应从0开始计算，否则在边界时无法计算。
3. 第二位数字。假设前面取余后为m，则第二位数字是第 m/(n-2)! 个未使用的数字。
4. 不断重复2，3，取余并且对(n-k)!进行除法，直至计算完毕

以下为主页君的代码，敬请指正：

解法1：

采用比较复杂的boolean来计算数字的索引（我们需要用一个boolean的数组来记录未使用的数字）：

 1 package Algorithms.permutation;
 2 
 3 /*
 4  The set [1,2,3,…,n] contains a total of n! unique permutations.
 5 
 6 By listing and labeling all of the permutations in order,
 7 We get the following sequence (ie, for n = 3):
 8 
 9     "123"
10     "132"
11     "213"
12     "231"
13     "312"
14     "321"
15 
16 Given n and k, return the kth permutation sequence.
17 
18 Note: Given n will be between 1 and 9 inclusive.
19  * */
20 public class PermutationSequence {
21     public static String getPermutation(int n, int k) {
22         if (n == 0) {
23             return "";
24         }
25         
26         // 先计算出(n)!
27         int num = 1;
28         for (int i = 1; i <= n; i++) {
29             num *= i;
30         }
31         
32         boolean[] use = new boolean[n];
33         for (int i = 0; i < n; i++) {
34             use[i] = false;
35         }
36         
37         // 因为index是从0开始计算 
38         k--;
39         StringBuilder sb = new StringBuilder();
40         for (int i = 0; i < n; i++) {
41             // 计算完第一个数字前，num要除以(n)
42             num = num / (n - i);
43             
44             int index = k / num;
45             k = k % num;
46             
47             for (int j = 0; j < n; j++) {                
48                 if (!use[j]) {
49                     if (index == 0) {
50                         // 记录下本次的结果.
51                         sb.append((j + 1) + "");
52                         use[j] = true;
53                         break;
54                     }
55                     
56                     // 遇到未使用过的数字，记录index
57                     index--;
58                 }
59             }
60         }
61         
62         return sb.toString();
63     }
64 
65     public static void main(String[] args) {
66         System.out.println(getPermutation(3, 5));
67     }
68 
69 }

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