[LeetCode] 717. 1-bit and 2-bit Characters

本文介绍了一种特殊的二进制字符串解码算法，该算法能够判断字符串结尾是否必定为单比特字符。通过分析字符串中特定模式，算法能确定最后一个字符是否一定由单一比特组成。

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.

这个题目思路因为如果为1的话后面接下来的bit就没用, 所以往后跳两格, 如果是0, 跳一格, 最后如果指在最后一个元素, 那么肯定是要一个元素单独, 否则就是可以不单独.

Code

class Solution:
    def 12bits(self, bits):
        i, length = 0, len(bits) -1
        while i < length:
            i += bits[i] + 1
        return i == length

转载于:https://www.cnblogs.com/Johnsonxiong/p/9479192.html