LWC 56：717. 1-bit and 2-bit Characters

LWC 56：717. 1-bit and 2-bit Characters

传送门：717. 1-bit and 2-bit Characters

Problem:

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.

思路：
如果遇到1，则可以把连带的[1,1]或者[1,0]取出来，如果遇到0，则只需要取一位，所以取到第n-2位，那么必然最后剩下一位或者零位，剩下一位返回true，剩下零位返回false.

代码如下：

    public boolean isOneBitCharacter(int[] bits) {        int n = bits.length;        int i = 0;        for (; i < n - 1;) {            if (bits[i] == 1) {                i += 2;            }            else {                i += 1;            }        }        return i == n - 1;    }