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.

题目比较暧昧= =细细理解下

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

之后看来一些更好的方法，可以把循环里面的二层嵌套的if提出来

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        
        int n = bits.size();
        
        int i = 0;
        
        for(;i < n - 1;)//最后一位一定为0
        {
            if(bits[i])
            {
                i += 2;
            }
            else
            {
                ++i;    
            }
        }
        
        return i == n - 1 ? true:false;
    }
};