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. 这题说数组中最后一个数字为0，问最后一个字符是不是one-bit-character。因为0就是one-bit-character,10是two-bit-character。所以判断最后一个数字0前面连续1有多少个，
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int len=bits.length;
if(len==1)
return true;
if(bits[len-2]==0)
return true;
else{
int k=1;//从倒数第二个字符开始，连续1的个数
int i=len-3;
while(i>=0){
if(bits[i]==1)
k++;
else
break;
i--;
}
if(k%2==0)
return true;
else
return false;
}
}
}