［LeetCode］1-bit and 2-bit Charac

链接：https://leetcode.com/problems/1-bit-and-2-bit-characters/description/
难度：Easy
题目：717. 1-bit and 2-bit 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.

翻译：给定两种字符，一种用1位的0表示，另一种用2位10或者11表示。现在给出一个由很多位字符表示的字符串，判断最后一位是不是1位字符。

思路：从头到尾遍历，如果该位数字为1，则向后前进两位，否则前进1位，循环的条件是i < n-1，即留出最后一位，所以当循环退出后，当i正好停留在n-1上，说明最后一位是单独分割开的。

参考代码：
Java

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