【刷题笔记】60. Permutation Sequence(改)

题目

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 and k, return the 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"

题目大意

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："123"，"132"，"213"，"231"，"312"，"321"，给定 n 和 k，返回第 k 个排列。

解题思路

• 用 DFS 暴力枚举，这种做法时间复杂度特别高，想想更优的解法。

参考代码

package leetcode

import (
	"fmt"
	"strconv"
)

func getPermutation(n int, k int) string {
	if k == 0 {
		return ""
	}
	used, p, res := make([]bool, n), []int{}, ""
	findPermutation(n, 0, &k, p, &res, &used)
	return res
}

func findPermutation(n, index int, k *int, p []int, res *string, used *[]bool) {
	fmt.Printf("n = %v index = %v k = %v p = %v res = %v user = %v\n", n, index, *k, p, *res, *used)
	if index == n {
		*k--
		if *k == 0 {
			for _, v := range p {
				*res += strconv.Itoa(v + 1)
			}
		}
		return
	}
	for i := 0; i < n; i++ {
		if !(*used)[i] {
			(*used)[i] = true
			p = append(p, i)
			findPermutation(n, index+1, k, p, res, used)
			p = p[:len(p)-1]
			(*used)[i] = false
		}
	}
	return
}