PROBLEM DESCRIPTION
Given the head of a singly linked list, return true if it is a palindrome. Palindrome LinkedList
SOLUTION
Using Stack (O(n) space complexity)
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class Solution {
public boolean isPalindrome(ListNode head) {
int length=0;
ListNode temp=head;
while(temp!=null){
length++;
temp=temp.next;
}
boolean skip = (length%2==1?true:false);
Stack s = new Stack();
temp=head;
for(int i=0;i<length;i++){
if(skip && i==(length/2)){
temp=temp.next;
continue;
}
if(i<(length/2)){
s.push(temp.val);
}else{
if(!s.empty()){
if((int)s.peek() == temp.val){
s.pop();
}else{
return false;
}
}
}
temp = temp.next;
}
return s.empty();
}
}
Optimized Approach (O(1) space complexity)
- Reverse 2nd half of the LinkedList
- Compare and check if elements match
- Reverse it back again so that we don’t modify the given LinkedList