717. 1-bit and 2-bit Characters 1 比特与 2 比特字符

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/1-bit-and-2-bit-characters/description/

题目描述

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

题目大意

有两种字符，一种是0，一种是10或者11，现在要判断整个数组是否由这两种组成的，要求最后一位的数字必须是单个的0．

解题方法

遍历

这个题真的很简单，因为有两种字符串，一种是0，一种是10或11。即一种长度是1，一种长度是2. 所以找个指针然后遍历一遍，看看当前值是0还是1，是0走1步，是1走两步。最后如果能到达len-1即可。

下面是第一次提交，想到了判断最后一个字符是不是0。题目中已经明确告诉了是0，所以下面有个改进版的。

"""
class Solution(object):
    def isOneBitCharacter(self, bits):
        """
        :type bits: List[int]
        :rtype: bool
        """
        pos = 0
        while pos < len(bits) - 1:
            if bits[pos] == 1:
                pos += 2
            else:
                pos += 1
        return pos == len(bits) - 1 and bits[pos] == 0

精简：

class Solution(object):
    def isOneBitCharacter(self, bits):
        """
        :type bits: List[int]
        :rtype: bool
        """
        pos = 0
        while pos < len(bits) - 1:
            pos += 2 if bits[pos] == 1 else 1
        return pos == len(bits) - 1

日期

2018 年 1 月 22 日 2018 年 11 月 14 日 —— 很严重的雾霾