# 排列序列

给出集合 [1,2,3,...,n],其所有元素共有 n! 种排列。

按大小顺序列出所有排列情况,并一一标记,当 n = 3 时, 所有排列如下:

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

给定 n 和 k,返回第 k 个排列。

 

示例 1:

输入:n = 3, k = 3
输出:
"213"

示例 2:

输入:n = 4, k = 9
输出:
"2314"

示例 3:

输入:n = 3, k = 1
输出:
"123"

 

提示:

## template ```python class Solution(object): def getPermutation(self, n, k): """ :type n: int :type k: int :rtype: str """ import math res=[""] def generate(s,k): n=len(s) if n<=2: if k==2: res[0]+=s[::-1] else: res[0]+=s return step = math.factorial(n-1) yu=k%step if yu==0: yu=step c=k//step-1 else: c=k//step res[0]+=s[c] generate(s[:c]+s[c+1:],yu) return s="" for i in range(1,n+1): s+=str(i) generate(s,k) return res[0] if __name__ == '__main__': s = Solution() print (s.getPermutation(3, 2)) ``` ## 答案 ```python ``` ## 选项 ### A ```python ``` ### B ```python ``` ### C ```python ```