[Swift]LeetCode60. 第k个排列 | Permutation Sequence

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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"

---

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：

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "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

class Solution {
    func getPermutation(_ num: Int, _ kth: Int) -> String {
        guard num > 1 else {
            return "1"
        }
        var vals: [Int] = Array(repeating: 1, count: 11)
        var nums: [Int] = Array(repeating: 1, count: 11)
        var kth = kth - 1
        for idx in 0...10 {
            vals[idx] = idx == 0 ? 1 : vals[idx - 1] * (idx + 1)
            nums[idx] = idx + 1
        }
        
        var result = ""
        for idx in (0...num - 2).reversed() {
            let iidx = kth / vals[idx]
            result.append(String(nums[iidx]))
            nums.remove(at: iidx)
            kth %= vals[idx]
        }
        result.append(String(nums[0]))
        return result
    }
}

---

12ms

class Solution {
    func getPermutation(_ n: Int, _ k: Int) -> String {
        guard n > 0, k > 0 else { return "" }
        if n-1 < 1 {
            return "(n)"
        }
        
        var arr: [Int] = []
        for i in 1...n {
            arr.append(i)
        }
        
        var factList: [Int] = []
        var current = 1
        for i in 1...n {
            current = current * i
            factList.append(current)
        }
        
        var output = ""
        for i in 1...n-1 {
            var idx = Int((Double((k % factList[n-i])*arr.count)/Double(factList[n-i])).rounded(.up))-1
            if idx < 0 {
                idx += arr.count
            }
            output.append("(arr.remove(at: idx))")
        }
        output.append("(arr.removeLast())")
        return output
    }
}

---

12ms

class Solution {
    func getPermutation(_ n: Int, _ k: Int) -> String {
        var k = k
        var j: Int
        var f: Int = 1
        var s = [Character](repeating: "0", count: n)
        let map: [Int: Character] = [1: "1", 2: "2", 3: "3", 4: "4", 5: "5", 6: "6", 7: "7", 8: "8", 9: "9"]
        for i in 1...n {
            f *= i
            s[i-1] = map[i]!
        }
        k -= 1
        for i in 0..<n {
            f /= n - i
            j = i + k / f
            let c = s[j]
            if j > i {
                for m in (i+1...j).reversed() {
                    s[m] = s[m-1]
                }
            }
            k %= f
            s[i] = c
        }
        return String(s)
    }
}

---

16ms

class Solution {
    func getPermutation(_ n: Int, _ k: Int) -> String {
        var nums: [Int] = []
        for i in 1...n {
            nums.append(i)
        }
        var result: String = ""
        backtracking(&nums, n, k-1, factorial(n - 1), &result)
        return result
    }
    func backtracking(_ nums: inout [Int], _ n: Int, _ k: Int, _ count: Int, _ result: inout String) {
        if nums.count == 1 {
            result += String(nums[0])
            return
        }
        
        var index = k / count
        var nextIndex = k % count
        
        
        result += String(nums.remove(at: index))
        backtracking(&nums, n - 1, nextIndex, count / (n - 1), &result)
    }
    func factorial(_ n: Int) -> Int {
        var n = n
        var result = 1
        while n > 1 {
            result *= n
            n -= 1
        }
        return result
    }
}

---

16ms

class Solution {
    func getPermutation(_ n: Int, _ k: Int) -> String {
        if n == 1 { return "1" }
        var array = [Int]()
        var dp = [Int](repeating: 1, count: n + 1)
        for i in 1 ... n {
            array.append(i)
            dp[i] = dp[i - 1] * i
        }
        
        var result = [Int]()
        var k = k
        while result.count < n {
            var i = 0
            while k > (i + 1) * dp[array.count - 1] {
                i += 1
            }
            k = k - i * dp[array.count - 1]
            let temp = array.remove(at: i)
            result.append(temp)
            
        }
        var s = ""
        for i in 0 ..< result.count {
            s += "(result[i])"
        }
        return s
    }
}