717. 1-bit and 2-bit Characters 1、2位字符

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

.
class Solution {
public boolean isOneBitCharacter(int[] bits) {
if(bits==null||bits[bits.length-1]!=0)return false;
int n=bits.length;
int i=0;
while(i<n-1){
if(bits[i]==0)i++;
else i=i+2;
}
return i==n-1;
}
}

题目解释：就是最后的一个必须是一位，只要遇到1就表示必然是两位挨着的，如11和10都符合2字符要求