检查字符串是否为回文的递归函数

给定一个字符串，编写一个递归函数来检查给定的字符串是否是回文，否则，不是回文。

例子：

Input : malayalam
Output : Yes
Reverse of malayalam is also
malayalam.

Input : max
Output : No
Reverse of max is not max. 

我们在这里讨论了一个迭代函数。
递归函数的想法很简单：

1) If there is only one character in string
   return true.
2) Else compare first and last characters
   and recur for remaining substring.

下面是上述思想的实现：

C++

// A recursive C++ program to
// check whether a given number
// is palindrome or not
#include 
using namespace std;
 
// A recursive function that
// check a str[s..e] is
// palindrome or not.
bool isPalRec(char str[],
              int s, int e)
{
     
    // If there is only one character
    if (s == e)
    return true;
 
    // If first and last
    // characters do not match
    if (str[s] != str[e])
    return false;
 
    // If there are more than
    // two characters, check if
    // middle substring is also
    // palindrome or not.
    if (s < e + 1)
    return isPalRec(str, s + 1, e - 1);
 
    return true;
}
 
bool isPalindrome(char str[])
{
    int n = strlen(str);
     
    // An empty string is
    // considered as palindrome
    if (n == 0)
        return true;
     
    return isPalRec(str, 0, n - 1);
}
 
// Driver Code
int main()
{
    char str[] = "geeg";
 
    if (isPalindrome(str))
    cout << "Yes";
    else
    cout << "No";
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

// A recursive C program to
// check whether a given number
// is palindrome or not
#include 
#include 
#include 
 
// A recursive function that
// check a str[s..e] is
// palindrome or not.
bool isPalRec(char str[],
              int s, int e)
{
    // If there is only one character
    if (s == e)
    return true;
 
    // If first and last
    // characters do not match
    if (str[s] != str[e])
    return false;
 
    // If there are more than
    // two characters, check if
    // middle substring is also
    // palindrome or not.
    if (s < e + 1)
    return isPalRec(str, s + 1, e - 1);
 
    return true;
}
 
bool isPalindrome(char str[])
{
int n = strlen(str);
 
// An empty string is
// considered as palindrome
if (n == 0)
    return true;
 
return isPalRec(str, 0, n - 1);
}
 
// Driver Code
int main()
{
    char str[] = "geeg";
 
    if (isPalindrome(str))
    printf("Yes");
    else
    printf("No");
 
    return 0;
}

Java

// A recursive JAVA program to
// check whether a given String
// is palindrome or not
import java.io.*;
 
class GFG
{
    // A recursive function that
    // check a str(s..e) is
    // palindrome or not.
    static boolean isPalRec(String str,
                            int s, int e)
    {
        // If there is only one character
        if (s == e)
            return true;
 
        // If first and last
        // characters do not match
        if ((str.charAt(s)) != (str.charAt(e)))
            return false;
 
        // If there are more than
        // two characters, check if
        // middle substring is also
        // palindrome or not.
        if (s < e + 1)
            return isPalRec(str, s + 1, e - 1);
 
        return true;
    }
 
    static boolean isPalindrome(String str)
    {
        int n = str.length();
 
    // An empty string is
    // considered as palindrome
        if (n == 0)
            return true;
 
        return isPalRec(str, 0, n - 1);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String str = "geeg";
 
        if (isPalindrome(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed
// by Nikita Tiwari

Python

# A recursive Python program
# to check whether a given
# number is palindrome or not
 
# A recursive function that
# check a str[s..e] is
# palindrome or not.
def isPalRec(st, s, e) :
     
    # If there is only one character
    if (s == e):
        return True
 
    # If first and last
    # characters do not match
    if (st[s] != st[e]) :
        return False
 
    # If there are more than
    # two characters, check if
    # middle substring is also
    # palindrome or not.
    if (s < e + 1) :
        return isPalRec(st, s + 1, e - 1);
 
    return True
 
def isPalindrome(st) :
    n = len(st)
     
    # An empty string is
    # considered as palindrome
    if (n == 0) :
        return True
     
    return isPalRec(st, 0, n - 1);
 
 
# Driver Code
st = "geeg"
if (isPalindrome(st)) :
    print "Yes"
else :
    print "No"
     
# This code is contributed
# by Nikita Tiwari.

C#

// A recursive C# program to
// check whether a given number
// is palindrome or not
using System;
 
class GFG
{
 
    // A recursive function that
    // check a str(s..e)
    // is palindrome or not.
    static bool isPalRec(String str,
                         int s,
                         int e)
    {
         
        // If there is only one character
        if (s == e)
            return true;
 
        // If first and last character
        // do not match
        if ((str[s]) != (str[e]))
            return false;
 
        // If there are more than two
        // characters, check if middle
        // substring is also
        // palindrome or not.
        if (s < e + 1)
            return isPalRec(str, s + 1,
                            e - 1);
             return true;
    }
 
    static bool isPalindrome(String str)
    {
        int n = str.Length;
 
        // An empty string is considered
        // as palindrome
        if (n == 0)
            return true;
 
        return isPalRec(str, 0, n - 1);
    }
 
    // Driver Code
    public static void Main()
    {
        String str = "geeg";
 
        if (isPalindrome(str))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed by Nitin Mittal.

Javascript

    // A recursive javascript program to
    // check whether a given String
    // is palindrome or not
    // A recursive function that
    // check a str(s..e) is
    // palindrome or not.
    function isPalRec( str , s , e) {
        // If there is only one character
        if (s == e)
            return true;
 
        // If first and last
        // characters do not match
        if ((str.charAt(s)) != (str.charAt(e)))
            return false;
 
        // If there are more than
        // two characters, check if
        // middle substring is also
        // palindrome or not.
        if (s < e + 1)
            return isPalRec(str, s + 1, e - 1);
 
        return true;
    }
 
    function isPalindrome( str) {
        var n = str.length;
 
        // An empty string is
        // considered as palindrome
        if (n == 0)
            return true;
 
        return isPalRec(str, 0, n - 1);
    }
 
    // Driver Code
     
        var str = "geeg";
 
        if (isPalindrome(str))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by gauravrajput1

输出

Yes

另一种方法：

基本上在遍历时检查第 i 个和第 ni-1 个索引是否相等。

如果不相等，则返回 false，如果它们相等，则继续递归调用。

C++

#include 
using namespace std;
 
bool isPalindrome(string s, int i){
        
    if(i > s.size()/2){
       return true ;
    }
    
    return s[i] == s[s.size()-i-1] && isPalindrome(s, i+1) ;
    
}
     
     
int main()
{
    string str = "geeg" ;
    if (isPalindrome(str, 0))
    cout << "Yes";
    else
    cout << "No";
 
    return 0;
 
}

输出

Yes

时间复杂度： O(n)

辅助空间： O(n)