LeetCode 1-bit and 2-bit Charact

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.

解法一（普通）：

    bool isOneBitCharacter(vector<int>& bits) {
        int n = bits.size();
        if(n == 1) return true;
        if(bits[n-2] == 0) return true;
        if(n == 2 || (n>=3 && bits[n-3] == 0))
            return false;
        //how many '1' continuous
        int i = n-2;
        while(i>=0 && bits[i] == 1) 
            i--;
        if((n-2-i) % 2 == 0) return true;
        else return false;
    }

解法二（高端）：

    bool isOneBitCharacter(vector<int>& bits) {
        int i = 0, n = bits.size();
        while(i < n-1){
            if(bits[i] == 0) i++;
            else i += 2;
        }
        return i == n-1;
    }

看到0后移一个，看到1后移两个。