LeetCode – Palindrome Linked List (Java)

Given a singly linked list, determine if it is a palindrome.

Java Solution 1 - Creat a new reversed list

We can create a new list in reversed order and then compare each node. The time and space are O(n).

public boolean isPalindrome(ListNode head) {
    if(head == null)
        return true;
 
    ListNode p = head;
    ListNode prev = new ListNode(head.val);
 
    while(p.next != null){
        ListNode temp = new ListNode(p.next.val);
        temp.next = prev;
        prev = temp;
        p = p.next;
    }
 
    ListNode p1 = head;
    ListNode p2 = prev;
 
    while(p1!=null){
        if(p1.val != p2.val)
            return false;
 
        p1 = p1.next;
        p2 = p2.next;
    }
 
    return true;
}

Java Solution 2 - Break and reverse second half

We can use a fast and slow pointer to get the center of the list, then reverse the second list and compare two sublists. The time is O(n) and space is O(1).

public boolean isPalindrome(ListNode head) {
 
    if(head == null || head.next==null)
        return true;
 
    //find list center
    ListNode fast = head;
    ListNode slow = head;
 
    while(fast.next!=null && fast.next.next!=null){
        fast = fast.next.next;
        slow = slow.next;
    }
 
    ListNode secondHead = slow.next;
    slow.next = null;
 
    //reverse second part of the list
    ListNode p1 = secondHead;
    ListNode p2 = p1.next;
 
    while(p1!=null && p2!=null){
        ListNode temp = p2.next;
        p2.next = p1;
        p1 = p2;
        p2 = temp;
    }
 
    secondHead.next = null;
 
    //compare two sublists now
    ListNode p = (p2==null?p1:p2);
    ListNode q = head;
    while(p!=null){
        if(p.val != q.val)
            return false;
 
        p = p.next;
        q = q.next;
 
    }
 
    return true;
}

Java Solution 3 - Recursive

public class Solution {
    ListNode left;
 
    public boolean isPalindrome(ListNode head) {
        left = head;
 
        boolean result = helper(head);
        return result;
    }
 
    public boolean helper(ListNode right){
 
        //stop recursion
        if (right == null)
            return true;
 
        //if sub-list is not palindrome,  return false
        boolean x = helper(right.next);
        if (!x)
            return false;
 
        //current left and right
        boolean y = (left.val == right.val);
 
        //move left to next
        left = left.next;
 
        return y;
    }
}

Time is O(n) and space is O(n).

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alexwest11

obv we can find middle, then unwind recurse , checking nodes
n/2-1 , n/2-2… checking w/ n/2 n/2+1 .. on

keep pointer to 2d half
alexwest11

can be re-reversed back.
alexwest11

it is not size. size can be calculated!,

how do you get j-1 from j!! you need traverse whole range.

it is o(n^2)
alexwest11

can use stack or array/queue, just push half elements, and compare 2d half in reverse.
Ustas51

Reversing half of the list destroys the original list.
Catherine

how to you know the linkedlist size and index if an unlimited listNode was provided
ivs

Recursive solution can’t be O(1) space right?
mg

public boolean isPalindrome(ListNode head) { if(head == null || head.next == null){ return true; } ListNode slow = head; ListNode fast = head;
while(fast.next != null && fast.next.next != null){ slow = slow.next; fast = fast.next.next; } ListNode reversed = reverseList(slow); slow.next = null; ListNode current = head; while(current != null){ if(reversed != null && current.val != reversed.val){ return false; } current = current.next; reversed = reversed.next; } return true; } ListNode reverseList(ListNode head){ if(head == null || head.next == null){ return head; } ListNode current = head; ListNode reversed = null; ListNode nextNode = null;
while(current != null){ nextNode = current.next; current.next = reversed; reversed = current; current = nextNode; } return reversed; }
baby

can we do something like this? the time is O(n) and space is O(1).

public static boolean isPalindrome(LinkedList list){
int i = 0;
int j = list.size() – 1;

while(i<j){
if (!list.get(i).equals(list.get(j))){
return false;
}
i++;
j–;
}
return true;
}
Gokhan A

public boolean isPalindromeLinkedList(Node head){
if(head == null){
throw new IllegalArgumentException();
}
if(head.next == null){
return true;
}
Node p1 = head;
Node p2 = head;
Node p = null;

while(p1 = null && p1.next != null){
p1 = p1.next.next;
p = p2;
p2 = p2.next;
}

Node prev = new Node(p2.value);

while(p2.next != null){
Node tmp = new Node(p2.next.value);
tmp.next = prev;
prev = tmp;
p2 = p2.next;
}
p1 = head;
p = p.next;

while(p.next != null){
if(p1 != p){
return false;
}
p1 = p1.next;
p = p.next;
}
return true;
}
Hooman

I know (;
WorstBuy

If you use stack, then the space is O(n) not O(1)
😉
Hooman

Using Stack:
private boolean isPalindromeUsingStack(ListNode head) { int count = 0; ListNode current = head; while (current!= null) { current = current.next; count++; }
current = head; // remove the middle node if the count is odd if (count % 2 == 1) { for (int i = 0; i < (count/2) - 1 ; i++) { current = current.next; } current.next = current.next.next; } // now head has even number of elements Stack stack = new Stack(); stack.push(head); current = head.next; while (current!= null) { if (current.val == stack.peek().val) { stack.pop(); } else { stack.push(current); } current = current.next; } return stack.isEmpty();
}
Truong Khanh Nguyen

Thanks for your code. I also discussion with illustration the problem here http://www.capacode.com/data-structure/linked-list/check-linked-list-palindrome/

LeetCode – Palindrome Linked List (Java)

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