不含回文的最长子字符串的长度 _回文最长子字符串的长度

本文概述

C ++
Java
Python3
C#
的PHP

给定一个小写字符串, 找到不包含回文的最长子字符串的长度作为子字符串。
例子：

Input : str = "daiict" Output : 3 dai, ict are longest substring that do not contain any palindrome as substringInput : str = "a" Output : 0 a is itself a palindrome

这样做的目的是观察如果任何字符形成回文, 则不能将其包含在任何子字符串中。因此, 在这种情况下, 将从构成回文的字符之前或之后选择所需的子字符串。
因此, 一种简单的解决方案是遍历字符串, 并针对每个字符检查其与相邻字符是否形成长度为2或3的回文。如果, 则不增加子字符串的长度, 否则将子字符串的长度重新初始化为零。使用这种方法可以找到最大子串的长度。
【不含回文的最长子字符串的长度】下面是上述方法的实现：
C ++

//C++ implementation of the above approach#include < bits/stdc++.h> using namespace std; //Function to find the length of the longest //substring int lenoflongestnonpalindrome(string s) { //initializing the variables int max1 = 1, len = 0; for ( int i = 0; i < s.length() - 1; i++) { //checking palindrome of size 2 //example: aa if (s[i] == s[i + 1]) len = 0; //checking palindrome of size 3 //example: aba else if (s[i + 1] == s[i - 1] & & i> 0) len = 1; else //incrementing length of substring len++; max1 = max(max1, len + 1); //finding maximum }//if there exits single character then //it is always palindrome if (max1 == 1) return 0; else return max1; }//Driver Code int main() { string s = "synapse" ; cout < < lenoflongestnonpalindrome(s) < < "\n" ; return 0; }

Java

//Java implementation of the above approach import java.util.Arrays; import java.lang.Math; class GFG {//Function to find the length of the longest //substring public static int lenoflongestnonpalindrome(String s) { //initializing the variables int max1 = 1 , len = 0 ; char [] new_str = s.toCharArray(); for ( int i = 0 ; i < new_str.length - 1 ; i++) { //checking palindrome of size 2 //example: aa if (new_str[i] == new_str[i + 1 ]) len = 0 ; //checking palindrome of size 3 //example: aba else if (i> 0 & & (new_str[i + 1 ] == new_str[i - 1 ])) len = 1 ; else //incrementing length of substring len++; max1 = Math.max(max1, len + 1 ); //finding maximum }//if there exits single character then //it is always palindrome if (max1 == 1 ) return 0 ; else return max1; }//Driver Code public static void main(String[] args) { String s = "synapse" ; System.out.println(lenoflongestnonpalindrome(s)); } }//This code is contributed by princiraj1992

Python3

# Python3 implementation of the above approach # Function to find the length # of the longest substring def lenoflongestnonpalindrome(s): # initializing the variables max1, length = 1 , 0for i in range ( 0 , len (s) - 1 ): # checking palindrome of # size 2 example: aa if s[i] = = s[i + 1 ]: length = 0# checking palindrome of # size 3 example: aba elif s[i + 1 ] = = s[i - 1 ] and i> 0 : length = 1 else : # incrementing length of substring length + = 1 max1 = max (max1, length + 1 ) # finding maximum # If there exits single character # then it is always palindrome if max1 = = 1 : return 0 else : return max1 # Driver Code if __name__ = = "__main__" :s = "synapse" print (lenoflongestnonpalindrome(s))# This code is contributed by Rituraj Jain

//C# implementation of the above approach using System; class GFG {//Function to find the length of the longest //substring public static int lenoflongestnonpalindrome(String s) { //initializing the variables int max1 = 1, len = 0; char [] new_str = s.ToCharArray(); for ( int i = 0; i < new_str.Length - 1; i++) { //checking palindrome of size 2 //example: aa if (new_str[i] == new_str[i + 1]) len = 0; //checking palindrome of size 3 //example: aba else if (i> 0 & & (new_str[i + 1] == new_str[i - 1])) len = 1; else //incrementing length of substring len++; max1 = Math.Max(max1, len + 1); //finding maximum }//if there exits single character then //it is always palindrome if (max1 == 1) return 0; else return max1; }//Driver Code public static void Main(String[] args) { String s = "synapse" ; Console.WriteLine(lenoflongestnonpalindrome(s)); } }//This code has been contributed by 29AjayKumar

的PHP

< ?php //PHP implementation of the above approach //Function to find the length of the longest //substring function lenoflongestnonpalindrome( $s ) { //initializing the variables $max1 = 1; $len = 0; for ( $i = 0; $i < strlen ( $s ) - 1; $i ++) { //checking palindrome of size 2 //example: aa if ( $s [ $i ] == $s [ $i + 1]) $len = 0; //checking palindrome of size 3 //example: aba else if ( $s [ $i + 1] == $s [ $i - 1] & & $i> 0) $len = 1; else //incrementing length of substring $len ++; $max1 = max( $max1 , $len + 1); //finding maximum } //if there exits single character then //it is always palindrome if ( $max1 == 1) return 0; else return $max1 ; } //Driver Code $s = "synapse" ; echo lenoflongestnonpalindrome( $s ), "\n" ; //This code is contributed by AnkitRai01?>

输出如下：