60.全排列

Permutation Sequence

问题描述：

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.

知识补充：

#divmod函数//本函数是实现a除以b，然后返回商与余数的元组。如果两个参数a,b都是整数，那么会采用整数除法，结果相当于（a//b, a % b)。如果a或b是浮点数，相当于（math.floor(a/b), a%b)。index, k = divmod(k, n)

参考答案（C++）：

class Solution {public:    string getPermutation(int n, int k) {        int i,j,f=1;        // left part of s is partially formed permutation, right part is the leftover chars.        string s(n,'0');        for(i=1;i<=n;i++){            f*=i;            s[i-1]+=i; // make s become 1234...n        }        for(i=0,k--;i<n;i++){            f/=n-i;            j=i+k/f; // calculate index of char to put at s[i]            char c=s[j];            // remove c by shifting to cover up (adjust the right part).            for(;j>i;j--)                s[j]=s[j-1];            k%=f;            s[i]=c;        }        return s;    }};

性能：

参考答案（python）：

import mathclass Solution:    # @param {integer} n    # @param {integer} k    # @return {string}    def getPermutation(self, n, k):        numbers = range(1, n+1)        permutation = ''        k -= 1        while n > 0:            n -= 1            # get the index of current digit            index, k = divmod(k, math.factorial(n))            permutation += str(numbers[index])            # remove handled number            numbers.remove(numbers[index])        return permutation

性能：