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.

在這道題中，給定了兩類特殊字符，一類是一位字符：0；一類是兩位字符：10或11.

現給了一個由0和1組成的、且最後一位是0的數組，判斷能否將其正確分割，且最後一位是單個字符0.

思路：如果出現1，則必須跟後面的0或1組成一個兩位字符。因此從前往後遍歷，如果bits[i] = 1，則將bits[i + 1]改為1；循環結束後，如果數組最後一位是1，則一定是一個兩位字符，如果是0，則是一個一位字符。具體代碼如下：

public class Solution {
    public boolean
 
 isOneBitCharacter(int[] bits){
        for(int i = 0; i < bits.length - 1; i++){
            if(bits[i] == 1){
                bits[i + 1] = 1;
                i++;
            }
        }
        return bits[bits.length - 1] == 0; 
    }

LeetCode 717. 1-bit and 2-bit Characters