Description:
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.

题意：一个数组只包含数字0和1，其中，数字(10,11)和(0)的组合分别会产生两种字符，要求按照这种组合，最后给定的数组能否得到最后的一个字符是由(0)这个组合产生的；

解法：其实，当我们在遍历数组时，只要判断当前的数字是1还是0就可以知道下一步了；如果是1的话，那么下一个字符只能和这个1产生组合，如果这个位置的数字是0的话，那么只能和自己产生组合，我们只要在遍历数组时按照这个规则去判断的话，一直到遍历到数组只剩下最后一个位置时，那么得到的结果就是true，否则就是false；

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        return check(bits, 0);
    }

    private boolean check(int[] bits, int pos) {
        if (pos == bits.length - 1 ) {
            return true;
        }
        if (pos + 1 < bits.length && bits[pos] == 1) {
            return check(bits, pos + 2);
        }
        if (pos < bits.length && bits[pos] == 0) {
            return check(bits, pos + 1);
        }
        return