从通过网络发送的数据包中解码消息

通过网络，消息作为数据包发送。每个数据包是一系列字节（8bits = 1 byte）。任务是解码长度为N的数据包 ( str )。数据包的格式如下：

• 数据包可以分成两部分。
• 第一个字节代表一个唯一值，比如K 。
• 数据包中剩余的字节代表要解码的消息。
• 每个消息字节都将与第一个字节进行异或运算，从而产生一组字母数字 ASCII字符以获取可读消息。

如果数据包的长度无效返回-1。

例子：

Input: str = “101010101110001011001111110001101100011011000101”, N = 48
Output: Hello
Explanation: Here K = 10101010, and the other bytes are:
“11100010”, “11001111”, “11000110”, “11000110”, “11000101”.
The Xor value with K are “01001000”, “01100101”, “01101100”, “01101100”, “01101111”
which have ASCII values 72, 101, 108, 108, 111 and the characters are ‘H’, ‘e’, ‘l’, ‘l’ and ‘o’.

Input: str = “1010101011100010111000101111”, N = 28
Output: -1
Explanation: The length of the packet is not valid. One byte of the message is missing some bits.

方法：问题的解决方案是基于执行。检查数据包长度的有效性。如果有效，则按照问题中的说明进行操作。请按照以下步骤操作：

• 检查输入长度是否可以被 8 整除，如果不是，则返回 -1。
• 现在，对于第一个字节，即前 8 位，计算十进制值并将其存储在一个变量中。
• 现在，对于每个即将到来的字节，将其与存储的第一个字节进行异或，并打印具有 ASCII 值的相应字符。

下面是上述方法的实现。

Java

// Java program to implement the approach
import java.util.*;
 
public class Main {
 
    // Function to decode the message
    static String decode(String str, int N)
    {
        // If length of packet is invalid
        if (N % 8 != 0)
            return "-1";
 
        String res = "";
        int j = 0, k = 0, first = 0, K = 0;
 
        for (int i = 0; i < N; i++) {
            k++;
 
            // If one byte length is reached
            if (k == 8) {
                k = 0;
                String s
                    = str.substring(j, i + 1);
                int n
                    = Integer.parseInt(s, 2);
                j = i + 1;
 
                // If not the first byte add
                // character to decoded message
                if (first == 1) {
                    int z = n ^ K;
                    char ch = (char)z;
                    res += ch;
                }
 
                // If it is the first byte
                else if (first == 0) {
                    first = 1;
                    K = n;
                }
            }
        }
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "101010101110001011001111110001101100011011000101";
        int N = 48;
        String ans = decode(str, N);
        System.out.println(ans);
    }
}

Python3

# Python code for the above approach
 
# Function to decode the message
def decode(strr: str, N: int) -> str:
     
    # If length of packet is invalid
    if N % 8 != 0:
        return "-1"
 
    res = ""
     
    j, k, first, K = 0, 0, 0, 0
     
    for i in range(N):
        k += 1
           
        # If one byte length is reached
        if k == 8:
            k = 0
            s = strr[j : i + 1]
            n = int(s, 2)
            j = i + 1
             
            # If not the first byte add
            # character to decoded message
            if first == 1:
                z = n ^ K
                ch = chr(z)
                res += ch
                 
            # If it is the first byte
            elif first == 0:
                first = 1
                K = n
    return res
   
# Driver code
if __name__ == "__main__":
     
    strr = "101010101110001011001111110001101100011011000101"
    N = 48
    ans = decode(strr, N)
    print(ans)
 
# This code is contributed by N SaiSuprabhanu

Javascript

        // JavaScript code for the above approach
 
        // Function to decode the message
        function decode(str, N)
        {
         
            // If length of packet is invalid
            if (N % 8 != 0)
                return "-1";
 
            let res = "";
            let j = 0, k = 0, first = 0, K = 0;
 
            for (let i = 0; i < N; i++) {
                k++;
 
                // If one byte length is reached
                if (k == 8) {
                    k = 0;
                    let s = str.slice(j, i + 1);
                    let n = parseInt(s, 2);
                    j = i + 1;
 
                    // If not the first byte add
                    // character to decoded message
                    if (first == 1) {
                        let z = n ^ K;
                        let ch = String.fromCharCode(z);
                        res += ch;
                    }
 
                    // If it is the first byte
                    else if (first == 0) {
                        first = 1;
                        K = n;
                    }
                }
            }
            return res;
        }
 
        // Driver code
        let str = "101010101110001011001111110001101100011011000101";
        let N = 48;
        let ans = decode(str, N);
        document.write(ans);
 
       // This code is contributed by Potta Lokesh
    

输出

Hello

时间复杂度： O(N)
辅助空间： O(logN)