[leetcode-717-1-bit and 2-bit Characters]

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：

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。

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int n = bits.size(), i = 0;
        while (i < n - 1) {
            if (bits[i] == 0) i++;
            else i += 2;
        }
        return i == n - 1 ? true : false;
    }
};

参考：

https://discuss.leetcode.com/topic/108743/java-solution-1-or-2