# 717. 1-bit and 2-bit Characters

**难度: Easy**

## 刷题内容

原题连接

内容描述

```
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.
```

思路 1

*- 时间复杂度: O(N)**- 空间复杂度: O(1)*

代码很好理解，直接看代码

```python

class Solution:

def isOneBitCharacter(self, bits: List[int]) -> bool:

def helper(bits): # 返回bits是否可以成功decode

if not bits:

return True

if len(bits) == 1:

if bits[0] == 0:

# 717. 1-bit and 2-bit Characters

**难度: Easy**

## 刷题内容

原题连接

内容描述

```
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.
```

思路 1

*- 时间复杂度: O(N)**- 空间复杂度: O(1)*

代码很好理解，直接看代码

```python

class Solution:

def isOneBitCharacter(self, bits: List[int]) -> bool:

def helper(bits): # 返回bits是否可以成功decode

if not bits:

return True

if len(bits) == 1:

if bits[0] == 0:

# 717. 1-bit and 2-bit Characters

**难度: Easy**

## 刷题内容

原题连接

内容描述

```
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.
```

思路 1

*- 时间复杂度: O(N)**- 空间复杂度: O(1)*

代码很好理解，直接看代码

```python

class Solution:

def isOneBitCharacter(self, bits: List[int]) -> bool:

def helper(bits): # 返回bits是否可以成功decode

if not bits:

return True

if len(bits) == 1:

if bits[0] == 0: