链表中最长的回文C ++？

Question

所以我想找出链表中是否存在回文，并打印该值。但是问题是即时通讯使用长整型值，并且想要在节点内甚至跨节点查找回文。即给定的列表：123-> 32166-> 78-> 1222最长的披肩应该是：123321我已经尝试了很多事情，但我似乎无法理解该怎么做。好的，如果您不显示任何代码，但是我很想听听解决此问题的任何想法。我有一些想法：将节点放入数组中，然后尝试在其中找到一个花环（尽管会占用大量内存）-反转列表并进行比较，但是存在每个节点中具有不同大小的整数值的问题即将78与32166进行比较将是不明智的-在尾部和头部分别创建两个指针，如果两个值都为true，则一个一个地移动，但这也有很多问题。

某些代码：

    int palindrome()
        {
    //  node *prev,*cur,*next;
       // cur=head;
      //    prev=next=NULL;
        list obj1;
        node *s,*e;
        s=obj1.head;
        e=head;
        while(s && e)
        {
        while(s->data==e->data)
        {
            e=e->next;
            s=s->next;
            cout<<"\n\n\n"<<e->data<<s->data;       
        }
        if(s->data!=e->data)
        {
            e=e->next;
        }
        }

        }

Answer 1

您不必使用LinkedList。试试这个解决方案，归功于GeeksforGeeks：

// A O(n^2) time and O(1) space program to  
// find the longest palindromic substring  
#include <bits/stdc++.h> 
using namespace std; 

// A utility function to print a substring str[low..high]  
void printSubStr(char* str, int low, int high)  
{  
    for( int i = low; i <= high; ++i )  
        cout << str[i];  
}  

// This function prints the longest palindrome substring (LPS)  
// of str[]. It also returns the length of the longest palindrome  
int longestPalSubstr(char *str)  
{  
    int maxLength = 1; // The result (length of LPS)
    int start = 0;  
    int len = strlen(str);    
    int low, high;  
    // One by one consider every character as center point of  
    // even and length palindromes  
    for (int i = 1; i < len; ++i)  
    {  
        // Find the longest even length palindrome  
        // with center points as i-1 and i.  
        low = i - 1;  
        high = i;  
        while (low >= 0 && high < len && str[low] == str[high])  
        {  
            if (high - low + 1 > maxLength)  
            {  
                start = low;  
                maxLength = high - low + 1;  
            }  
            --low;  
            ++high;  
        }  

        // Find the longest odd length palindrome with center  
        // point as i  
        low = i - 1;  
        high = i + 1;  
        while (low >= 0 && high < len && str[low] == str[high])  
        {  
            if (high - low + 1 > maxLength)  
            {  
                start = low;  
                maxLength = high - low + 1;  
            }  
            --low;  
            ++high;  
        }  
    }  

    cout<<"Longest palindrome substring is: ";  
    printSubStr(str, start, start + maxLength - 1);  

    return maxLength;  
}  

// Driver program to test above functions  
int main()  
{  
    char str[] = "forgeeksskeegfor";  
    cout<<"\nLength is: "<<longestPalSubstr(str)<<endl;  
    return 0;  
}