将中间字符串解码为原始字符串
给定一个以中间形式写的字符串s，将其改回原始字符串。字符串字符串的字母。如果字符串的长度是偶数，则中间字母是两个中间字母的左边。给定的字符串是通过写下单词的中间字母，然后将其删除并重复该过程直到没有字母留下来形成的。
例子：
Input: eekgs
Output: geeks
Explanation: in the original string “geeks”
can be written in median form by picking up
e first then, again e, then k then g and at
the end s. As these are the median when the
median letter is picked and deleted.
Input: abc
Output: bac
Explanation: median of bac is a, then median
of bc is b, then median of c is c. 为了找到答案，我们可以从左到右遍历给定的编码字符串，并添加答案字符串中的每个字母，一个字母到开头，下一个字母到结尾，下一个字母开始，等等。如果 n 是偶数，则必须将第一个字母添加到开头，将第二个字母添加到末尾。在另一种情况下，第一个字母到结尾，第二个到开头。我们需要做到这一点，直到我们不添加给定字符串中的所有字母。
注意：对于偶数长度的字符串，当我们将第一个字符添加到开头并将第二个字符添加到末尾时，剩余的字符串将始终是偶数长度。对于奇数长度的字符串也是如此。
下面给出的是上述方法的实现
C++
// C++ program to decode a median string
// to the original string
#include
using namespace std;
// function to calculate the median back string
string decodeMedianString(string s)
{
// length of string
int l = s.length();
// initialize a blank string
string s1 = "";
// Flag to check if length is even or odd
bool isEven = (l % 2 == 0)? true : false;
// traverse from first to last
for (int i = 0; i < l; i += 2) {
// if len is even then add first character
// to beginning of new string and second
// character to end
if (isEven) {
s1 = s[i] + s1;
s1 += s[i + 1];
} else {
// if current length is odd and is
// greater than 1
if (l - i > 1) {
// add first character to end and
// second character to beginning
s1 += s[i];
s1 = s[i + 1] + s1;
} else {
// if length is 1, add character
// to end
s1 += s[i];
}
}
}
return s1;
}
// driver program
int main()
{
string s = "eekgs";
cout << decodeMedianString(s);
return 0;
} Java
// java program to decode a median
// string to the original string
public class GFG {
// function to calculate the
// median back string
static String decodeMedianString(String s)
{
// length of string
int l = s.length();
// initialize a blank string
String s1 = "";
// Flag to check if length is
// even or odd
boolean isEven = (l % 2 == 0) ?
true : false;
// traverse from first to last
for (int i = 0; i < l; i += 2)
{
// if len is even then add
// first character to
// beginning of new string
// and second character to
// end
if (isEven) {
s1 = s.charAt(i) + s1;
s1 += s.charAt(i+1);
}
else {
// if current length is
// odd and is greater
// than 1
if (l - i > 1) {
// add first character
// to end and second
// character to
// beginning
s1 += s.charAt(i);
s1 = s.charAt(i+1) + s1;
}
else {
// if length is 1,
// add character
// to end
s1 += s.charAt(i);
}
}
}
return s1;
}
// Driver code
public static void main(String args[])
{
String s = "eekgs";
System.out.println(
decodeMedianString(s));
}
}
// This code is contributed by Sam007.Python3
# Python3 program to decode a median
# string to the original string
# function to calculate the median
# back string
def decodeMedianString(s):
# length of string
l = len(s)
# initialize a blank string
s1 = ""
# Flag to check if length is
# even or odd
if(l % 2 == 0):
isEven = True
else:
isEven = False
# traverse from first to last
for i in range(0, l, 2):
# if len is even then add first
# character to beginning of new
# string and second character to end
if (isEven):
s1 = s[i] + s1
s1 += s[i + 1]
else :
# if current length is odd and
# is greater than 1
if (l - i > 1):
# add first character to end and
# second character to beginning
s1 += s[i]
s1 = s[i + 1] + s1
else:
# if length is 1, add character
# to end
s1 += s[i]
return s1
# Driver Code
if __name__ == '__main__':
s = "eekgs"
print(decodeMedianString(s))
# This code is contributed by
# Sanjit_PrasadC#
// C# program to decode a median
// string to the original string
using System;
class GFG {
// function to calculate the
// median back string
static string decodeMedianString(string s)
{
// length of string
int l = s.Length;
// initialize a blank string
string s1 = "";
// Flag to check if length is
// even or odd
bool isEven = (l % 2 == 0) ?
true : false;
// traverse from first to last
for (int i = 0; i < l; i += 2)
{
// if len is even then add
// first character to
// beginning of new string
// and second character to
// end
if (isEven) {
s1 = s[i] + s1;
s1 += s[i + 1];
}
else {
// if current length is
// odd and is greater
// than 1
if (l - i > 1) {
// add first character
// to end and second
// character to
// beginning
s1 += s[i];
s1 = s[i + 1] + s1;
}
else {
// if length is 1,
// add character
// to end
s1 += s[i];
}
}
}
return s1;
}
// Driver code
public static void Main ()
{
string s = "eekgs";
Console.WriteLine(
decodeMedianString(s));
}
}
// This code is contributed by Sam007.Javascript
输出：
geeks