解码使用给定算法编码的字符串

给定一个使用以下编码算法解码的解码字符串str ：
记下字符串的中间字符，然后将其删除并重复该过程，直到没有字符为止。例如， “abba”将被编码为“bbaa” 。
注意，当字符串的长度为偶数时，中间字符是两个中间字符的第一个字符。
例子：

Input: "ofrsgkeeeekgs"
Output: geeksforgeeks

Input: str = "bbaa"
Output: abba

的方法：可以观察到的是，虽然字符串进行解码，该编码字符串的首字母成为解码的字符串的中值。因此，首先，写入编码字符串的第一个字符并将其从编码字符串删除，然后开始将编码字符串的第一个字符首先添加到解码字符串的左侧，然后添加到解码字符串的右侧，并重复执行此任务直到编码字符串变为空。
例如：

Encoded String          Decoded String
ofrsgkeeeekgs           o
frsgkeeeekgs            fo
rsgkeeeekgs             for
sgkeeeekgs              sfor
gkeeeekgs               sforg
keeeekgs                ksforg
eeeekgs                 ksforge
eeekgs                  eksforge
eekgs                   eksforgee
ekgs                    eeksforgee
kgs                     eeksforgeek
gs                      geeksgorgeek
s                       geeksforgeeks                     

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to decode and print
// the original string
void decodeStr(string str, int len)
{
 
    // To store the decoded string
    char c[len] = "";
    int med, pos = 1, k;
 
    // Getting the mid element
    if (len % 2 == 1)
        med = len / 2;
    else
        med = len / 2 - 1;
 
    // Storing the first element of the
    // string at the median position
    c[med] = str[0];
 
    // If the length is even then store
    // the second element also
    if (len % 2 == 0)
        c[med + 1] = str[1];
 
    // k represents the number of characters
    // that are already stored in the c[]
    if (len & 1)
        k = 1;
    else
        k = 2;
 
    for (int i = k; i < len; i += 2) {
        c[med - pos] = str[i];
 
        // If string length is odd
        if (len % 2 == 1)
            c[med + pos] = str[i + 1];
 
        // If it is even
        else
            c[med + pos + 1] = str[i + 1];
        pos++;
    }
 
    // Print the decoded string
    for (int i = 0; i < len; i++)
        cout << c[i];
}
 
// Driver code
int main()
{
    string str = "ofrsgkeeeekgs";
    int len = str.length();
 
    decodeStr(str, len);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG{
 
// Function to decode and print
// the original String
static void decodeStr(String str, int len)
{
 
    // To store the decoded String
    char []c = new char[len];
    int med, pos = 1, k;
 
    // Getting the mid element
    if (len % 2 == 1)
        med = len / 2;
    else
        med = len / 2 - 1;
 
    // Storing the first element of the
    // String at the median position
    c[med] = str.charAt(0);
 
    // If the length is even then store
    // the second element also
    if (len % 2 == 0)
        c[med + 1] = str.charAt(1);
 
    // k represents the number of characters
    // that are already stored in the c[]
    if (len % 2 == 1)
        k = 1;
    else
        k = 2;
 
    for(int i = k; i < len; i += 2)
    {
       c[med - pos] = str.charAt(i);
        
       // If String length is odd
       if (len % 2 == 1)
           c[med + pos] = str.charAt(i + 1);
            
       // If it is even
       else
           c[med + pos + 1] = str.charAt(i + 1);
       pos++;
    }
 
    // Print the decoded String
    for (int i = 0; i < len; i++)
        System.out.print(c[i]);
}
 
// Driver code
public static void main(String[] args)
{
    String str = "ofrsgkeeeekgs";
    int len = str.length();
 
    decodeStr(str, len);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the
# above approach
 
# Function to decode and print
# the original string
def decodeStr(str, len):
 
    # To store the decoded string
    c = ["" for i in range(len)]
    pos = 1
 
    # Getting the mid element
    if(len % 2 == 1):
        med = int(len / 2)
    else:
        med = int(len / 2 - 1)
 
    # Storing the first element 
    # of the string at the
    # median position
    c[med] = str[0]
 
    # If the length is even
    # then store the second
    # element also
    if(len % 2 == 0):
        c[med + 1] = str[1]
 
    # k represents the number
    # of characters that are
    # already stored in the c[]
    if(len & 1):
        k = 1
    else:
        k = 2
 
    for i in range(k, len, 2):
        c[med - pos] = str[i]
 
        # If string length is odd
        if(len % 2 == 1):
            c[med + pos] = str[i + 1]
 
        # If it is even
        else:
            c[med + pos + 1] = str[i + 1]
        pos += 1
 
    # Print the decoded string
    print(*c, sep = "")
 
# Driver code
str = "ofrsgkeeeekgs"
len = len(str)
decodeStr(str, len)
 
# This code is contributed by avanitrachhadiya2155

C#

// C# implementation of the approach
using System;
 
class GFG{
 
// Function to decode and print
// the original String
static void decodeStr(String str, int len)
{
 
    // To store the decoded String
    char []c = new char[len];
     
    int med, pos = 1, k;
 
    // Getting the mid element
    if (len % 2 == 1)
        med = len / 2;
    else
        med = len / 2 - 1;
 
    // Storing the first element of the
    // String at the median position
    c[med] = str[0];
 
    // If the length is even then store
    // the second element also
    if (len % 2 == 0)
        c[med + 1] = str[1];
 
    // k represents the number of characters
    // that are already stored in the c[]
    if (len % 2 == 1)
        k = 1;
    else
        k = 2;
 
    for(int i = k; i < len; i += 2)
    {
       c[med - pos] = str[i];
        
       // If String length is odd
       if (len % 2 == 1)
           c[med + pos] = str[i + 1];
        
       // If it is even
       else
           c[med + pos + 1] = str[i + 1];
       pos++;
    }
 
    // Print the decoded String
    for(int i = 0; i < len; i++)
       Console.Write(c[i]);
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "ofrsgkeeeekgs";
    int len = str.Length;
 
    decodeStr(str, len);
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
 
// JavaScript implementation of the approach
 
// Function to decode and print
// the original string
function decodeStr(str, len)
{
 
    // To store the decoded string
    var c = Array(len).fill("");
    var med, pos = 1, k;
 
    // Getting the mid element
    if (len % 2 == 1)
        med = parseInt(len / 2);
    else
        med = parseInt(len / 2) - 1;
 
    // Storing the first element of the
    // string at the median position
    c[med] = str[0];
 
    // If the length is even then store
    // the second element also
    if (len % 2 == 0)
        c[med + 1] = str[1];
 
    // k represents the number of characters
    // that are already stored in the c[]
    if (len & 1)
        k = 1;
    else
        k = 2;
 
    for (var i = k; i < len; i += 2) {
        c[med - pos] = str[i];
 
        // If string length is odd
        if (len % 2 == 1)
            c[med + pos] = str[i + 1];
 
        // If it is even
        else
            c[med + pos + 1] = str[i + 1];
        pos++;
    }
 
    // Print the decoded string
    for (var i = 0; i < len; i++)
    {
        document.write(c[i]);
    }
}
 
// Driver code
var str = "ofrsgkeeeekgs";
var len = str.length;
decodeStr(str, len);
 

输出：

geeksforgeeks

复杂度： O(n)

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