717. 1-bit and 2-bit Characters--easy

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.

1.思考

• 根据题意可知，遍历整个序列，若当前是0，则index++；若当前是1，则index = index + 2；
• 直到序列最后两位，若倒数第二位为1，则index = index + 2，最后一位不是1bit，return false；
• 若倒数第二位为0，则看最后一位是否为0，是则return true，否则return false；
• 即当解码跳出循环时为indexlen-1，且bits[index]0，return true；否则，return false。
  
  

2.实现
Runtime: 4ms(100%)
Memory: 8.5MB(100%)

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