https://www.freecodecamp.org/news/everything-you-need-to-know-about-cookies-for-web-development/
- Method 1
Basically we can sole the palindrome problem in two way
first is by using the reminder concept means that we can find the number using modulo operator and reverse the number and if given number is equal to that number then we can say number is palindrome.
if(x<0) return false;
if(x==0) return true;
if(x%10==x) return true;
if(x/10==0) return false;
while(x>0)
{
rem=x%10;
n=n*10+rem;
x=x/10;
}
if(n==y)
return true;
else
return false;
}
Method 2:
This method is used when number is given using the Linked List or array form or vector form
For example number is given in the linked list then we can create a vector v and store the value in vector and check from left pointing value and from right and if at least one number which is not equal the return false and if not then return true after the termination of loop
Algo 2 :
ListNode *temp=head;
vector<int> v;
while(temp!=NULL)
{
v.push_back(temp->val);
temp=temp->next;
}
int n=v.size()-1;
for(int i=0;i<v.size()/2;i++)
{
if(v[i]!=v[n])
return false;
n--;
}
return true;
Thanks for reading
Happy Coding
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