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

.
分析：

class Solution {
public boolean isOneBitCharacter(int[] bits) {
//给定数组，最后一位是0，其中规定1bit只能是0，2bit是10或者11。返回该数组能只有这两种bit组成
//思路：分两种情况，2bit为10和11，1bit为0。那么只需要统计2bit出现的次数和1bit出现的次数
//最后将次数乘以对应的长度和总的数组长度对比即可

int even=0,odd=0;
for(int i=0;i<bits.length;i++){
if(i<bits.length-1&&bits[i]==1&&(bits[i+1]==0||bits[i+1]==1)){
//为2bit的情况
i++;
even++;
}else if(i<bits.length-1&&bits[i]==0){
odd++;
}

if(i==bits.length-1&&bits[i]==0){
odd++;
}
}

return even*2+odd==bits.length?true:false;

}
}