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 always0
or1
.
题目大致意思是现在有两种特殊字符一个是0一个是10或者11,给一个数组,最后一位一定是0,问最后一个是否只能用0编码
思路如下,从数组头开始扫描,如果是1则跳过下一个因为1一定和下一个组成一个字符,如果是0则扫描下一个。
当正好落在最后一个元素时,返回true否则返回false
public boolean isOneBitCharacter(int[] bits) { int length = bits.length; int i=0; while(i<length) { if(i==(length-1)) { return true; } if(bits[i]==1) { i+=2; } else { i++; } } return false; }