leetcode 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.

直接判断就好了。对于当前位，是1的话直接条到i+2是0的话不用管，然后看最后一位是不是0

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