LeetCode 717. 1-bit and 2-bit Characters

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.

在这道题中，给定了两类特殊字符，一类是一位字符：0；一类是两位字符：10或11.

现给了一个由0和1组成的、且最后一位是0的数组，判断能否将其正确分割，且最后一位是单个字符0.

思路：如果出现1，则必须跟后面的0或1组成一个两位字符。因此从前往后遍历，如果bits[i] = 1，则将bits[i + 1]改为1；循环结束后，如果数组最后一位是1，则一定是一个两位字符，如果是0，则是一个一位字符。具体代码如下：

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; 
    }