717. 1-bit and 2-bit Characters

问题描述

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.

题目链接：

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思路分析

三种编码方式：0，10，11。问一串编码的最后一个字符是不是只有一位。

用一个bool表示当前位置的编码是2位还是1位。因为有1的话必定是两位的编码，所以直接+2，并置true。看循环到结束时以何种方式结束的。

代码

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

时间复杂度：$O(n)$
空间复杂度：$O(1)$

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反思

没什么难度，想明白题目要求就行了。