717. 1-bit and 2-bit Characters

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。

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        if(bits.size()==0) return false;
        if(bits.size()==1) {if(bits[0]!=0) return false;else return true;}
        int n =bits.size();
        int count_1=0;
        if(bits[n-1] != 0) return false;
        else{
            if(bits[n-2]==0) return true;
            else if(bits[n-2]!=0){
                   for(int i=n-2;i>=0;i--){
                       if(bits[i]!=0) count_1++;
                       else break;
                   }
            }
        }
        if(count_1%2==0) return true;
        else return false;

    }
};