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:
"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.
- 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
}