Tech Base

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.

數值只會是 0 或 1，而 1 bit character 是 0，2 bit character 則是 10 或 11，而最後一位數一定會是 0

所以只要數值為 1，則代表該數值加上他的下一個數值會是一個 2 bit character，所以建立一個指標，遇到數值為 1 就直接跳過下一個數值，如果最後 index 位置是指到最後一個數值，則代表最後一個剛好是 1 bit character，如果超過則代表最後會是 2 bit character

public boolean isOneBitCharacter(int[] bits) {
    int index = 0;
    while(index < bits.length - 1) {
        if(bits[index] == 1) {
            index += 2;
        } else {
            index += 1;
        }
    }

    return bits.length - 1 == index;
}