LeetCode 717. 1-bit and 2-bit Characters

题目：

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.

思路： 从第一位开始判断，如果它为1，那肯定是两位bit的字条，即11或者10，i加2；如果为0，那么肯定为1位的字符，i加1。

代码：

class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
int len = bits.size();
bool res;
for (int i = 0; i < len;){
if (bits[i]){//如果当前位为1，那么判断之后两位的，因为如果为1只能为11或者10
i += 2;
res = false;//结果置false
}
else{
i += 1;//如果当前位为0，那么肯定为单个0，结果置为true;
res = true;
}
}
return res;
}
};

结果： 7ms