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: TrueExplanation: 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: FalseExplanation: 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.

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