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; } } }