题目：
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：从bits的第一位开始，如果第一个数为1，则根据编码规则，第二个数字只能是1或0，则可以跳到第三个分析，如果是1的话可以直接跳到第五个数字，但是如果是0的话就要跳到第四个分析.
代码：Python
class Solution: def isOneBitCharacter(self, bits): """ :type bits: List[int] :rtype: bool """ i = 0 while i < len(bits): if i == len(bits)-1: if bits[i] == 0: return True else: return False else: if bits[i] == 1: i += 2 else: i += 1 return False