问题
有两种特殊字符。第一种字符可以用一比特0来表示。第二种字符可以用两比特(10 或 11)来表示。
现给一个由若干比特组成的字符串。问最后一个字符是否必定为一个一比特字符。给定的字符串总是由0结束。
输入: bits = [1, 0, 0]
输出: True
解释: 唯一的编码方式是一个两比特字符和一个一比特字符。所以最后一个字符是一比特字符。
输入: bits = [1, 1, 1, 0]
输出: False
解释: 唯一的编码方式是两比特字符和两比特字符。所以最后一个字符不是一比特字符。
注意:
1 <= len(bits) <= 1000.
bits[i] 总是0 或 1.
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.
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.
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.
示例
public class Program {
public static void Main(string[] args) {
int[] nums = null;
nums = new int[] { 1, 1, 1, 0 };
var res = IsOneBitCharacter(nums);
Console.WriteLine(res);
Console.ReadKey();
}
private static bool IsOneBitCharacter(int[] bits) {
//1总是要和后面的1个数字编码,即+2,0不用编码,+1往后继续即可
//这道题做不出来主动面壁思过吧
int i = 0;
while(i < bits.Length - 1) {
i = bits[i] == 0 ? ++i : i + 2;
}
return i == bits.Length - 1;
}
}
以上给出1种算法实现,以下是这个案例的输出结果:
False
分析:
显而易见,以上算法的时间复杂度为: 。