LeetCode Algorithm 0060 - Permutation Sequence (Medium)

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Problem Link: https://leetcode.com/problems/permutation-sequence/

Related Topics: Math Backtracking

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Description

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 for n = 3:

• 1, "123"

• 2, "132"

• 3, "213"

• 4, "231"

• 5, "312"

• 6, "321"

Given $n$n and $k$k , return the $k^{th}$kth permutation sequence.

Note:

• Given n will be between 1 and 9 inclusive.

• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

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Analysis

已知集合 $N$N 中有 $n$n 个数字 $\{1,2,...,n\}${1,2,...,n} 一共有 $n!$n! 种序列。即，排列 $P_n^n$Pnn​ 种序列。则：

• 第一位有 $P_n^1$Pn1​ 种序列，其它位有 $P_{n-1}^{n-1}$Pn−1n−1​ 种序列。即，每个数字有 $P_{n-1}^{n-1}$Pn−1n−1​ 种序列。那么第 $k$k 个序列第一位数字的位置 $i_1$i1​ 为
  $i_1 = \lfloor (k-1) \div P_{n-1}^{n-1} \rfloor$i1​=⌊(k−1)÷Pn−1n−1​⌋
  那么 取出 第一位数字 $n_1$n1​ 为
  $n_1 = N \left[ i_1 \right], \quad \text{取出后：} N \cap \{ n_1 \} = \phi$n1​=N[i1​],取出后：N∩{n1​}=ϕ
  即，在第一位数字从 $1$1 到 $n-1$n−1 共有 $n_1-1$n1​−1 个 $P_{n-1}^{n-1}$Pn−1n−1​ 序列。
  之后到达第 $k$k 个序列还需要 $k_1$k1​ 个序列，即
  $k-1 = k_1 (mod \; P_{n-1}^{n-1})$k−1=k1​(modPn−1n−1​)

• 同理，第二位数字的位置 $i_2$i2​ 为
  $i_2 = \lfloor k_1 \div P_{n-2}^{n-2} \rfloor$i2​=⌊k1​÷Pn−2n−2​⌋
  到达第 $k$k 个序列还需要 $k_2$k2​ 个序列，即
  $k_1 = k_2 (mod \; P_{n-2}^{n-2})$k1​=k2​(modPn−2n−2​)

• 最终，第 $i$i 位数字的位置 $i$i 为
  $i = \lfloor k_{i-1} \div P_{n-i}^{n-i} \rfloor$i=⌊ki−1​÷Pn−in−i​⌋
  到达第 $k$k 个序列还需要 $k_i$ki​ 个序列，即
  $k_{i-1} = k_i (mod \; P_{n-i}^{n-i})$ki−1​=ki​(modPn−in−i​)

举例 $n=5, k=8$n=5,k=8 ，则：

num	$i$i	$k-1=7$k−1=7	$N = \{ 1, 2, 3, 4, 5 \}$N={1,2,3,4,5}	$n=5$n=5
1	$i_1 = \lfloor 7 \div 4! \rfloor = 0$i1​=⌊7÷4!⌋=0	$k_1 = 7 \, mod \, 4! = 7$k1​=7mod4!=7	$N = \{ 1, 2, 3, 4, 5 \}$N={1,2,3,4,5}	$n_1 = N[0] = 1$n1​=N[0]=1
2	$i_2 = \lfloor 7 \div 3! \rfloor = 1$i2​=⌊7÷3!⌋=1	$k_2 = 7 \, mod \, 3! = 1$k2​=7mod3!=1	$N = \{ 2, 3, 4, 5 \}$N={2,3,4,5}	$n_2 = N[1] = 3$n2​=N[1]=3
3	$i_3 = \lfloor 1 \div 2! \rfloor = 0$i3​=⌊1÷2!⌋=0	$k_3 = 1 \, mod \, 2! = 1$k3​=1mod2!=1	$N = \{ 2, 4, 5 \}$N={2,4,5}	$n_3 = N[0] = 2$n3​=N[0]=2
4	$i_4 = \lfloor 1 \div 1! \rfloor = 1$i4​=⌊1÷1!⌋=1	$k_4 = 1 \, mod \, 1! = 0$k4​=1mod1!=0	$N = \{ 4, 5 \}$N={4,5}	$n_4 = N[1] = 5$n4​=N[1]=5
5	$i_5 = \lfloor 0 \div 0! \rfloor = 0$i5​=⌊0÷0!⌋=0	$k_5 = 0 \, mod \, 0! = 0$k5​=0mod0!=0	$N = \{ 4 \}$N={4}	$n_5 = N[0] = 4$n5​=N[0]=4

则最终结果为13254。

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Solution C++

// Author: https://blog.csdn.net/DarkRabbit
// Problem: https://leetcode.com/problems/permutation-sequence/
// Difficulty: Medium
// Related Topics: `Math` `Backtracking`

#pragma once

#include "pch.h"

namespace P60PermutationSequence
{
    class Solution
    {
        public:
        string getPermutation(int n, int k)
        {
            // 一共 n! 种序列
            // 当第一位确定后，后面共有 (n-1)! 种序列
            // 即，n个数有 n! = (n-1)! * n 种序列
            // 则，第k个序列的第一位为 k / (n-1)! + 1
            // 第二位是第 k % (n-1)! 个序列
            // 然后求第二位 (n-2)!，以此类推

            vector<int> nums;
            for (int i = 1; i <= n; i++)
            {
                nums.push_back(i);
            }

            int fac = 1; // 求n-1的阶乘
            for (int i = 2; i < n; i++)
            {
                fac *= i;
            }

            k--; // 数组从0开始

            string seq;
            int index;
            for (int i = n - 1; i >= 0; i--)
            {
                index = k / fac; // k / (n-1)! + 1，下标和数值差1
                seq.push_back(nums[index] + '0');
                nums.erase(nums.begin() + index); // 删除已经使用的数字

                k %= fac; // 求[n-1, n-2, ..., 1]个序列中的位置
                if (i > 0)
                {
                    fac /= i; // 求[(n-2)!, (n-3)!, ..., 0!]
                }
            }

            return seq;
        }

    };
}

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