标签：count return LeetCode717 比特 character bit bits
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 always0or1.
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
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 }
标签：count,return,LeetCode717,比特,character,bit,bits 来源： https://www.cnblogs.com/strengthen/p/10509872.html
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