[Swift]LeetCode717. 1比特與2比特字符 | 1-bit and 2-bit Characters

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We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).

Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.

Example 1:

Input: 
bits = [1, 0, 0]
Output: True
Explanation: 
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.

Example 2:

Input: 
bits = [1, 1, 1, 0]
Output: False
Explanation: 
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.

Note:

• 1 <= len(bits) <= 1000.
• bits[i] is always 0 or 1.

---

有兩種特殊字符。第一種字符能夠用一比特0來表示。第二種字符能夠用兩比特(10 或 11)來表示。

現給一個由若干比特組成的字符串。問最後一個字符是否一定爲一個一比特字符。給定的字符串老是由0結束。

示例 1:

輸入: 
bits = [1, 0, 0]
輸出: True
解釋: 
惟一的編碼方式是一個兩比特字符和一個一比特字符。因此最後一個字符是一比特字符。

示例 2:

輸入: 
bits = [1, 1, 1, 0]
輸出: False
解釋: 
惟一的編碼方式是兩比特字符和兩比特字符。因此最後一個字符不是一比特字符。

注意:

• 1 <= len(bits) <= 1000.
• bits[i] 老是0 或 1.

---

20ms

 1 class Solution 
 2 {    
 3     func isOneBitCharacter( _ bits: [ Int ] ) -> Bool 
 4     {        
 5         func findR1( _ startIndex: Int ) -> Bool
 6         {
 7             if startIndex == ( bits.count )
 8             {
 9                 return false
10             }
11             if startIndex == ( bits.count - 1 )
12             {
13                 return bits[ ( bits.count - 1 ) ] == 0
14             }            
15             var result: Bool            
16             if bits[ startIndex ] == 0
17             {                
18                 result = findR1( startIndex + 1 )                
19             }
20             else
21             {
22                 result = findR1( startIndex + 2 )
23             }            
24             return result            
25         }        
26         return findR1( 0 )
27     }    
28 }

---

Runtime: 24 ms

Memory Usage: 18.7 MB

 1 class Solution {
 2     func isOneBitCharacter(_ bits: [Int]) -> Bool {
 3         var n:Int = bits.count
 4         var i:Int = 0
 5         while (i < n - 1)
 6         {
 7             i += bits[i] + 1
 8         }
 9         return i == n - 1
10     }
11 }

---

24ms

 1 class Solution {
 2     func isOneBitCharacter(_ bits: [Int]) -> Bool {
 3         var skipping = false, is1bit = false
 4         
 5         for b in bits {
 6             if skipping {
 7                 skipping = false
 8                 continue
 9             }
10             if b == 1 {
11                 skipping = true
12                 is1bit = false
13             } else {
14                 is1bit = true
15             }
16         }
17         return is1bit
18     }
19 }

---

28ms

 1 class Solution {
 2     func isOneBitCharacter(_ bits: [Int]) -> Bool {
 3       var state = false
 4       var i = 0
 5       while i < bits.count {
 6           if bits[i] == 1{
 7               state = false
 8               i += 2
 9           }else{
10               state = true
11               i += 1
12           }
13       }
14       return state
15     }
16 }

---

52ms

 1 class Solution {
 2     func isOneBitCharacter(_ bits: [Int]) -> Bool {
 3         let count = bits.count
 4         guard count > 1 else { return true }
 5         
 6         var cursor = count - 2
 7         var oneCount = 0
 8         while cursor >= 0 {
 9             defer { cursor -= 1 }
10             if bits[cursor] == 1 {
11                 oneCount += 1
12             }
13             else {
14                 break
15             }
16         }
17         
18         return oneCount & 1 == 0
19     }
20 }

---

60ms

 1 class Solution {
 2     func isOneBitCharacter(_ bits: [Int]) -> Bool {
 3         var isTwoBit = false
 4         
 5         bits.enumerated().forEach { index, bit in
 6             if isTwoBit {
 7                 isTwoBit = (bits.count - 1 == index)
 8                 return
 9             }
10                                    
11             if bit == 1 {
12                 isTwoBit = true
13                 return
14             }
15         }
16         
17         return !isTwoBit
18     }
19 }