[Swift]LeetCode717. 1比特与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.

---

有两种特殊字符。第一种字符可以用一比特0来表示。第二种字符可以用两比特(10 或 11)来表示。

现给一个由若干比特组成的字符串。问最后一个字符是否必定为一个一比特字符。给定的字符串总是由0结束。

示例 1:

输入: 
bits = [1, 0, 0]
输出: True
解释: 
唯一的编码方式是一个两比特字符和一个一比特字符。所以最后一个字符是一比特字符。

示例 2:

输入: 
bits = [1, 1, 1, 0]
输出: False
解释: 
唯一的编码方式是两比特字符和两比特字符。所以最后一个字符不是一比特字符。

注意:

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

---

20ms

class Solution 
{    
    func isOneBitCharacter( _ bits: [ Int ] ) -> Bool 
    {        
        func findR1( _ startIndex: Int ) -> Bool
        {
            if startIndex == ( bits.count )
            {
                return false
            }
            if startIndex == ( bits.count - 1 )
            {
                return bits[ ( bits.count - 1 ) ] == 0
            }            
            var result: Bool            
            if bits[ startIndex ] == 0
            {                
                result = findR1( startIndex + 1 )                
            }
            else
            {
                result = findR1( startIndex + 2 )
            }            
            return result            
        }        
        return findR1( 0 )
    }    
}

---

Runtime: 24 ms

Memory Usage: 18.7 MB

class Solution {
    func isOneBitCharacter(_ bits: [Int]) -> Bool {
        var n:Int = bits.count
        var i:Int = 0
        while (i < n - 1)
        {
            i += bits[i] + 1
        }
        return i == n - 1
    }
}

---

24ms

class Solution {
    func isOneBitCharacter(_ bits: [Int]) -> Bool {
        var skipping = false, is1bit = false
        
        for b in bits {
            if skipping {
                skipping = false
                continue
            }
            if b == 1 {
                skipping = true
                is1bit = false
            } else {
                is1bit = true
            }
        }
        return is1bit
    }
}

---

28ms

class Solution {
    func isOneBitCharacter(_ bits: [Int]) -> Bool {
      var state = false
      var i = 0
      while i < bits.count {
          if bits[i] == 1{
              state = false
              i += 2
          }else{
              state = true
              i += 1
          }
      }
      return state
    }
}

---

52ms

class Solution {
    func isOneBitCharacter(_ bits: [Int]) -> Bool {
        let count = bits.count
        guard count > 1 else { return true }
        
        var cursor = count - 2
        var oneCount = 0
        while cursor >= 0 {
            defer { cursor -= 1 }
            if bits[cursor] == 1 {
                oneCount += 1
            }
            else {
                break
            }
        }
        
        return oneCount & 1 == 0
    }
}

---

60ms

class Solution {
    func isOneBitCharacter(_ bits: [Int]) -> Bool {
        var isTwoBit = false
        
        bits.enumerated().forEach { index, bit in
            if isTwoBit {
                isTwoBit = (bits.count - 1 == index)
                return
            }
                                   
            if bit == 1 {
                isTwoBit = true
                return
            }
        }
        
        return !isTwoBit
    }
}