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：

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)

反思

没什么难度，想明白题目要求就行了。