检查所有回文子串是否都是奇数长度

给定一个字符串's' 检查它的所有回文子字符串是否都是奇数长度。如果是，则打印“YES”或“NO”，否则打印。

例子：

Input: str = “geeksforgeeks” 
Output: NO 
Since, “ee” is a palindromic sub-string of even length.
Input: str = “madamimadam” 
Output: YES 

蛮力方法：

简单地说，遍历 's' 的每个子字符串并检查它是否是回文。如果它是一个回文，那么它必须是奇数长度。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to check if
// the string is palindrome
bool checkPalindrome(string s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s[i] != s[s.length() - i - 1])
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
bool CheckOdd(string s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{
     
    // Creating each substring
    string x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];
         
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }
     
    return true;
}
 
// Driver code
int main()
{
    string s = "geeksforgeeks";
    if(CheckOdd(s))
        cout<<("YES");
    else
        cout<<("NO");
}
// This code is contributed by
// Sahil_shelangia

Java

// Java implementation of the approach
import java.util.*;
class GFG
{
 
// Function to check if
// the string is palindrome
static boolean checkPalindrome(String s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s.charAt(i) != s.charAt(s.length() - i - 1))
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static boolean CheckOdd(String s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{
     
    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s.charAt(j);
         
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }
     
    return true;
}
 
// Driver code
public static void main(String args[])
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}
 
// This code is contributed
// by Arnab Kundu

Python

# Python implementation of the approach
 
# Function to check if
# the string is palindrome
 
 
def checkPalindrome(s):
    for i in range(len(s)):
        if(s[i] != s[len(s)-i-1]):
            return False
    return True
 
# Function that checks whether
# all the palindromic
# sub-strings are of odd length.
 
 
def CheckOdd(s):
    n = len(s)
    for i in range(n):
 
        # Creating each substring
        x = ""
        for j in range(i, n):
            x += s[j]
            # If the sub-string is
            # of even length and
            # is a palindrome then,
            # we return False
            if(len(x) % 2 == 0
                    and checkPalindrome(x) == True):
                return False
    return True
 
 
# Driver code
s = "geeksforgeeks"
if(CheckOdd(s)):
    print("YES")
else:
    print("NO")

C#

// C# implementation of the approach
using System;
                     
 
public class GFG {
 
// Function to check if
// the string is palindrome
static bool checkPalindrome(String s)
{
    for (int i = 0; i < s.Length; i++)
    {
        if(s[i] != s[(s.Length - i - 1)])
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static bool CheckOdd(String s)
{
int n = s.Length;
for (int i = 0; i < n; i++)
{
     
    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];
         
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.Length % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }
     
    return true;
}
 
// Driver code
public static void Main()
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}
 
/* This code is contributed by 29AjayKumar*/

Javascript

 
// JavaScript implementation of the approach
 
// Function to check if
// the string is palindrome
function checkPalindrome(s)
{
    for (let i = 0; i < s.length; i++)
    {
        if(s[i] != s[s.length - i - 1])
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
function CheckOdd(s)
{
    let n = s.length;
for (let i = 0; i < n; i++)
{
      
    // Creating each substring
    let x = "";
    for (let j = i; j < n; j++)
    {
        x += s[j];
          
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }
      
    return true;
}
 
// Driver code
let s = "geeksforgeeks";
if(CheckOdd(s))
    document.write("YES");
else
    document.write("NO");
 
 
// This code is contributed by avanitrachhadiya2155
 

输出

NO

有效方法：要检查 s 的所有回文子串是否都具有奇数长度，我们可以搜索它的偶数长度回文子串。我们知道每个偶数长度的回文至少有两个相同的连续字符（例如 cxxa、ee）。因此，我们可以一次检查两个连续的字符，看它们是否相同。如果是这样，那么 s 有一个偶数长度的回文子串，因此输出将为 NO，如果我们发现没有偶数长度的子串，答案将为 YES。

我们可以在一个字符串遍历之后完成这个检查。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function that checks whether s
// contains a even length palindromic
// sub-strings or not.
bool CheckEven(string s)
{
  for (int i = 1; i < s.size(); ++i) {
        if (s[i] == s[i - 1]) {
            return true;
        }
    }
    return false;
}
 
// Driver code
int main()
{
    string s = "geeksforgeeks";
    if(CheckEven(s)==false)
        cout<<("YES");
    else
        cout<<("NO");
}
// This code is contributed by
// Aditya Jaiswal

Java

// Java implementation of the approach
import java.util.*;
class GFG
{
 
// Function to check if
// the string is palindrome
static boolean checkPalindrome(String s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s.charAt(i) != s.charAt(s.length() - i - 1))
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static boolean CheckOdd(String s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{
     
    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s.charAt(j);
         
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }
     
    return true;
}
 
// Driver code
public static void main(String args[])
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Python implementation of the approach
 
# Function to check if
# the string is palindrome
def checkPalindrome(s):
    for i in range(len(s)):
        if(s[i] != s[len(s)-i-1]):
            return False
    return True
 
# Function that checks whether
# all the palindromic
# sub-strings are of odd length.
def CheckOdd(s):
    n = len(s)
    for i in range(n):
 
        # Creating each substring
        x = ""
        for j in range(i, n):
            x += s[j]
            # If the sub-string is
            # of even length and
            # is a palindrome then,
            # we return False
            if(len(x)% 2 == 0
                  and checkPalindrome(x) == True):
                return False
    return True
 
# Driver code
s = "geeksforgeeks"
if(CheckOdd(s)):
    print("YES")
else:
    print("NO")

C＃

// C# implementation of the approach
using System;
                     
 
public class GFG {
 
// Function to check if
// the string is palindrome
static bool checkPalindrome(String s)
{
    for (int i = 0; i < s.Length; i++)
    {
        if(s[i] != s[(s.Length - i - 1)])
            return false;
    }
    return true;
}
  
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static bool CheckOdd(String s)
{
int n = s.Length;
for (int i = 0; i < n; i++)
{
      
    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];
          
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.Length % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }
      
    return true;
}
  
// Driver code
public static void Main()
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}
  
/* This code is contributed by 29AjayKumar*/

Javascript

// Javascript implementation of the approach
 
// Function to check if
// the string is palindrome
function  checkPalindrome(s)
{
    for (let i = 0; i < s.length; i++)
    {
        if(s[i] != s[(s.length - i - 1)])
            return false;
    }
    return true;
}
 
// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
function CheckOdd(s)
{
    let n = s.length;
for (let i = 0; i < n; i++)
{
       
    // Creating each substring
    let x = "";
    for (let j = i; j < n; j++)
    {
        x += s[j];
           
        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }
       
    return true;
}
 
// Driver code
let s = "geeksforgeeks";
if(CheckOdd(s))
    document.write("YES");
else
    document.write("NO");
 
 
// This code is contributed by rag2127

输出：

NO

时间复杂度： O(N)
空间复杂度： O(1)