[Swift]LeetCode779. 第K个语法符号 | K-th Symbol in Grammar

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On the first row, we write a 0. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.

Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).

Examples:
Input: N = 1, K = 1
Output: 0

Input: N = 2, K = 1
Output: 0

Input: N = 2, K = 2
Output: 1

Input: N = 4, K = 5
Output: 1

Explanation:
row 1: 0
row 2: 01
row 3: 0110
row 4: 01101001

Note:

1. N will be an integer in the range [1, 30].
2. K will be an integer in the range [1, 2^(N-1)].

---

在第一行我们写上一个 0。接下来的每一行，将前一行中的0替换为01，1替换为10。

给定行数 N 和序数 K，返回第 N 行中第 K个字符。（K从1开始）

例子:

输入: N = 1, K = 1
输出: 0

输入: N = 2, K = 1
输出: 0

输入: N = 2, K = 2
输出: 1

输入: N = 4, K = 5
输出: 1

解释:
第一行: 0
第二行: 01
第三行: 0110
第四行: 01101001

注意：

1. N 的范围 [1, 30].
2. K 的范围 [1, 2^(N-1)].

---

Runtime: 4 ms

Memory Usage: 18.4 MB

class Solution {
    func kthGrammar(_ N: Int, _ K: Int) -> Int {
               if N == 1{
            return 0
        }
        
        return K % 2 == 0 ? (kthGrammar(N-1,K/2) == 0 ? 1 : 0 ):(kthGrammar(N-1,(K+1)/2) == 1 ? 1 : 0)
    }
}

---

Runtime: 4 ms

Memory Usage: 18.4 MB

class Solution {
    func kthGrammar(_ N: Int, _ K: Int) -> Int {
        var K = K
        var res:Int = 0
        while (K > 1)
        {
            K = (K % 2 == 1) ? K + 1 : K / 2
            res ^= 1
        }
        return res
    }
}

---

4ms

class Solution {
    func kthGrammar(_ N: Int, _ K: Int) -> Int {
        var s = pow(Double(2), Double(N - 1))
        var flips = 0
        var K = K
        while (s > 2){
            if K > Int(s / 2) {
                K -= Int(s / 2)
                flips += 1
            }
            s /= 2
        }
        K -= 1 // K is either 2 or 1
        if flips % 2 == 1 {
            K = 1 - K
        }
        return K        
    }
}

---

8ms

class Solution {
    func kthGrammar(_ N: Int, _ K: Int) -> Int {
        if K == 1 {
            return 0
        }
        let halfCount = Int(pow(2.0, Double(N-2)))
        let rm = K%halfCount
        if K <= halfCount {
            return kthGrammar(N-1, rm == 0 ? halfCount : rm)    
        } else {
            return kthGrammar(N-1, rm == 0 ? halfCount : rm) == 0 ? 1 : 0
        }        
    }
}