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Permutation Sequence

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.

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

分析

首先可以想到一个简单直白的方法,即调用 k-1next_permutation(),从而得到第k个排列。这个方法把前k个排列全部求出来了,比较浪费,时间复杂度是 O(kn),所以会超时。有没有办法直接求第k个排列呢?有!

利用康托编码的思路,假设有n个不重复的元素,第k个排列是a1,a2,a3,...,ana_1, a_2, a_3, ..., a_n,那么a1a_1是哪一个位置呢?

我们把a1a_1去掉,那么剩下的排列为 a2,a3,...,ana_2, a_3, ..., a_n, 共计n-1个元素,n-1个元素共有(n-1)!个排列,于是就可以知道 a1=k/(n1)!a_1 = k / (n-1)!

同理,a2,a3,...,ana_2, a_3, ..., a_n 的值推导如下:

k2=k%(n1)!k_2 = k\%(n-1)!

a2=k2/(n2)!a_2 = k_2/(n-2)!

\quad \cdots

kn1=kn2%2!k_{n-1} = k_{n-2}\%2!

an1=kn1/1!a_{n-1} = k_{n-1}/1!

an=0a_n = 0

康托编码

// Permutation Sequence
// 康托编码
// 时间复杂度O(n),空间复杂度O(1)
public class Solution {
public String getPermutation(int n, int k) {
string s(n, '0');
string result;
for (int i = 0; i < n; ++i)
s[i] += i + 1;

return kth_permutation(s, k);
}
private:
int factorial(int n) {
int result = 1;
for (int i = 1; i < n+1; ++i)
result *= i;
return result;
}

// s 已排好序,是第一个排列
string kth_permutation(string &s, int k) {
const int n = s.size();
string result;

int base = factorial(n - 1);
--k; // 康托编码从0开始

for (int i = n - 1; i > 0; k %= base, base /= i, --i) {
auto a = next(s.begin(), k / base);
result.push_back(*a);
s.erase(a);
}

result.push_back(s[0]); // 最后一个
return result;
}
};

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