LeetCode 717. 1-bit and 2-bit Characters

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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; 
    }

 

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