weekly contest 56 第一题 1-bit and 2-bit Characters

题目

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.

分析

题目的意思是说看这个串是否可以由两个字符“10”“11”或者一个字符“0”组成并且最后是一个字符0结尾

我们先把最后一位0弹出（题目已经说最后一位就是0）

然后判断前面是否可以由两个字符或一个字符组成

观察发现用1开头的一定是两个字符的

用这种贪心的思想

代码

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

时间复杂度O(N)

空间复杂度O(1)