Leetcode 717: 1-bit and 2-bit Characters

标签：log   ati   last   bsp   always   decode   pre   ring   immediate   

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 }

Leetcode 717: 1-bit and 2-bit Characters

标签：log   ati   last   bsp   always   decode   pre   ring   immediate   

原文地址：http://www.cnblogs.com/liangmou/p/7782893.html