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"
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。
说明：
- 给定 n 的范围是 [1, 9]。
- 给定 k 的范围是[1, n!]。
示例 1:
输入: n = 3, k = 3 输出: "213"
示例 2:
输入: n = 4, k = 9 输出: "2314"
8ms
1 class Solution {
2 func getPermutation(_ num: Int, _ kth: Int) -> String {
3 guard num > 1 else {
4 return "1"
5 }
6 var vals: [Int] = Array(repeating: 1, count: 11)
7 var nums: [Int] = Array(repeating: 1, count: 11)
8 var kth = kth - 1
9 for idx in 0...10 {
10 vals[idx] = idx == 0 ? 1 : vals[idx - 1] * (idx + 1)
11 nums[idx] = idx + 1
12 }
13
14 var result = ""
15 for idx in (0...num - 2).reversed() {
16 let iidx = kth / vals[idx]
17 result.append(String(nums[iidx]))
18 nums.remove(at: iidx)
19 kth %= vals[idx]
20 }
21 result.append(String(nums[0]))
22 return result
23 }
24 }
12ms
1 class Solution {
2 func getPermutation(_ n: Int, _ k: Int) -> String {
3 guard n > 0, k > 0 else { return "" }
4 if n-1 < 1 {
5 return "\(n)"
6 }
7
8 var arr: [Int] = []
9 for i in 1...n {
10 arr.append(i)
11 }
12
13 var factList: [Int] = []
14 var current = 1
15 for i in 1...n {
16 current = current * i
17 factList.append(current)
18 }
19
20 var output = ""
21 for i in 1...n-1 {
22 var idx = Int((Double((k % factList[n-i])*arr.count)/Double(factList[n-i])).rounded(.up))-1
23 if idx < 0 {
24 idx += arr.count
25 }
26 output.append("\(arr.remove(at: idx))")
27 }
28 output.append("\(arr.removeLast())")
29 return output
30 }
31 }
12ms
1 class Solution {
2 func getPermutation(_ n: Int, _ k: Int) -> String {
3 var k = k
4 var j: Int
5 var f: Int = 1
6 var s = [Character](repeating: "0", count: n)
7 let map: [Int: Character] = [1: "1", 2: "2", 3: "3", 4: "4", 5: "5", 6: "6", 7: "7", 8: "8", 9: "9"]
8 for i in 1...n {
9 f *= i
10 s[i-1] = map[i]!
11 }
12 k -= 1
13 for i in 0..<n {
14 f /= n - i
15 j = i + k / f
16 let c = s[j]
17 if j > i {
18 for m in (i+1...j).reversed() {
19 s[m] = s[m-1]
20 }
21 }
22 k %= f
23 s[i] = c
24 }
25 return String(s)
26 }
27 }
16ms
1 class Solution {
2 func getPermutation(_ n: Int, _ k: Int) -> String {
3 var nums: [Int] = []
4 for i in 1...n {
5 nums.append(i)
6 }
7 var result: String = ""
8 backtracking(&nums, n, k-1, factorial(n - 1), &result)
9 return result
10 }
11 func backtracking(_ nums: inout [Int], _ n: Int, _ k: Int, _ count: Int, _ result: inout String) {
12 if nums.count == 1 {
13 result += String(nums[0])
14 return
15 }
16
17 var index = k / count
18 var nextIndex = k % count
19
20
21 result += String(nums.remove(at: index))
22 backtracking(&nums, n - 1, nextIndex, count / (n - 1), &result)
23 }
24 func factorial(_ n: Int) -> Int {
25 var n = n
26 var result = 1
27 while n > 1 {
28 result *= n
29 n -= 1
30 }
31 return result
32 }
33 }
16ms
1 class Solution {
2 func getPermutation(_ n: Int, _ k: Int) -> String {
3 if n == 1 { return "1" }
4 var array = [Int]()
5 var dp = [Int](repeating: 1, count: n + 1)
6 for i in 1 ... n {
7 array.append(i)
8 dp[i] = dp[i - 1] * i
9 }
10
11 var result = [Int]()
12 var k = k
13 while result.count < n {
14 var i = 0
15 while k > (i + 1) * dp[array.count - 1] {
16 i += 1
17 }
18 k = k - i * dp[array.count - 1]
19 let temp = array.remove(at: i)
20 result.append(temp)
21
22 }
23 var s = ""
24 for i in 0 ..< result.count {
25 s += "\(result[i])"
26 }
27 return s
28 }
29 }
原文地址：https://www.cnblogs.com/strengthen/p/9920849.html
时间： 2024-10-21 23:42:04