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

.

题目大意：
一个数组仅由0和1组成，且最后一位是0.问整个数组是否仅可以由10,11,0这三种组成。
思路：
遍历数组，如果遇到1就让i加2，初始时设置i为0，遇到0就让i加1.如果最后i恰好等于bits.size()-1，就返回true，否则返回false。
注意while循环i的界限。
代码：
class Solution {
public:
bool isOneBitCharacter(vector<int>& bits) {
int i=0;
if(bits.size()<2)
return true;
while(i<bits.size()-1)
{
if(bits[i]==0)
i++;
else if(bits[i]==1)
i+=2;
}
if(i!=bits.size()-1)
return false;
else
return true;
}
};