0060. Permutation Sequence

60. Permutation Sequence #

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。

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
}


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