[LeetCode] Permutation Sequence, Solution_permutation sequence solution leetcode-优快云博客

本文介绍了一种通过数学方法解决排列问题的技术，用于找出给定数字集合中特定位置的排列序列。通过逐步减小问题规模，算法能够高效地确定序列中的每一个元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

» Solve this problem

[Thoughts]
两个解法。
第一，DFS
递归遍历所有可能，然后累加计算，直至到K为止。

第二，数学解法。

假设有n个元素，第K个permutation是
a1, a2, a3, ….. …, an
那么a1是哪一个数字呢？

那么这里，我们把a1去掉，那么剩下的permutation为
a2, a3, …. …. an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K
a1 = K1 / (n-1)!

同理，a2的值可以推导为
a2 = K2 / (n-2)!
K2 = K1 % (n-1)!
…….
a(n-1) = K(n-1) / 1!
K(n-1) = K(n-2) /2!

an = K(n-1)

实现如下：

1:       string getPermutation(int n, int k) {  
2:            vector<int> nums(n);  
3:            int permCount =1;  
4:            for(int i =0; i< n; i++)  
5:            {  
6:                 nums[i] = i+1;  
7:                 permCount *= (i+1);        
8:            }  
9:            // change K from (1,n) to (0, n-1) to accord to index  
10:            k--;  
11:            string targetNum;  
12:            for(int i =0; i< n; i++)  
13:            {    
14:                 permCount = permCount/ (n-i);  
15:                 int choosed = k / permCount;  
16:                 targetNum.push_back(nums[choosed] + '0');  
17:                 //restruct nums since one num has been picked  
18:                 for(int j =choosed; j< n-i; j++)  
19:                 {  
20:                      nums[j]=nums[j+1];  
21:                 }  
22:                 k = k%permCount;  
23:            }  
24:            return targetNum;  
25:       }