Leetcode 717: 1-bit and 2-bit Characters

最新推荐文章于 2022-11-10 02:07:59 发布

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最新推荐文章于 2022-11-10 02:07:59 发布

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原文链接：http://www.cnblogs.com/liangmou/p/7782893.html

本文介绍了一种算法，用于判断给定的比特串是否以一个一比特特殊字符结尾。通过解析比特串，该算法能够确定最后一个字符是否为规定的一比特字符。

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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.

 1 public class Solution {
 2     public bool IsOneBitCharacter(int[] bits) {        
 3         // Optimization: if the last bit is not 0, return false immediately
 4         if (bits[bits.Length - 1] != 0) return false;
 5         
 6         int i = 0;
 7         while (i < bits.Length)
 8         {
 9             if (bits[i] == 1)
10             {
11                 i += 2;
12             }
13             else if (i == bits.Length - 1)
14             {
15                 return true;
16             }
17             else
18             {
19                 i++;
20             }                
21         }
22         
23         return false;
24     }
25     
26 }