Problem description:

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.

Solution:

The idea is to determine whether if the last 0 is a 0 itself or belong to a 10. We can use a pointer to walk through the array. If the bits[i] == 1 then we know we will have a two-bit character. On the other hand, if it’s a 0, then we just increment 1.

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