717. 1-bit and 2-bit Characters

Posted 2020-10-30 一日一更

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.

 

解题思路：

先判断最后一位是0还是1.

然后看倒数第二位，如果是0那么肯定是，

如果是1的话，那么倒数第三位必需是1，并且连续的1必需是偶数。

1. class Solution {  
2. public:  
3.     bool isOneBitCharacter(vector<int>& bits) {  
4.         if(bits.size()==0) return false;  
5.         if(bits.size()==1) {if(bits[0]!=0) return false;else return true;}  
6.         int n =bits.size();  
7.         int count_1=0;  
8.         if(bits[n-1] != 0) return false;  
9.         else{  
10.             if(bits[n-2]==0) return true;  
11.             else if(bits[n-2]!=0){  
12.                    for(int i=n-2;i>=0;i--){  
13.                        if(bits[i]!=0) count_1++;  
14.                        else break;  
15.                    }  
16.             }  
17.         }  
18.         if(count_1%2==0) return true;  
19.         else return false;  
20.           
21.     }  
22. };