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

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

1 class Solution {
2     func kthGrammar(_ N: Int, _ K: Int) -> Int {
3                if N == 1{
4             return 0
5         }
6         
7         return K % 2 == 0 ? (kthGrammar(N-1,K/2) == 0 ? 1 : 0 ):(kthGrammar(N-1,(K+1)/2) == 1 ? 1 : 0)
8     }
9 }

---

Runtime: 4 ms

Memory Usage: 18.4 MB

 1 class Solution {
 2     func kthGrammar(_ N: Int, _ K: Int) -> Int {
 3         var K = K
 4         var res:Int = 0
 5         while (K > 1)
 6         {
 7             K = (K % 2 == 1) ? K + 1 : K / 2
 8             res ^= 1
 9         }
10         return res
11     }
12 }

---

4ms

 1 class Solution {
 2     func kthGrammar(_ N: Int, _ K: Int) -> Int {
 3         var s = pow(Double(2), Double(N - 1))
 4         var flips = 0
 5         var K = K
 6         while (s > 2){
 7             if K > Int(s / 2) {
 8                 K -= Int(s / 2)
 9                 flips += 1
10             }
11             s /= 2
12         }
13         K -= 1 // K is either 2 or 1
14         if flips % 2 == 1 {
15             K = 1 - K
16         }
17         return K        
18     }
19 }

---

8ms

 1 class Solution {
 2     func kthGrammar(_ N: Int, _ K: Int) -> Int {
 3         if K == 1 {
 4             return 0
 5         }
 6         let halfCount = Int(pow(2.0, Double(N-2)))
 7         let rm = K%halfCount
 8         if K <= halfCount {
 9             return kthGrammar(N-1, rm == 0 ? halfCount : rm)    
10         } else {
11             return kthGrammar(N-1, rm == 0 ? halfCount : rm) == 0 ? 1 : 0
12         }        
13     }
14 }