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ZeeshanAli-0704
ZeeshanAli-0704

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Delete the Middle Node of a Linked List

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list.

The middle node of a linked list of size n is the ⌊n / 2βŒ‹th node from the start using 0-based indexing, where ⌊xβŒ‹ denotes the largest integer less than or equal to x.

For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively.

Example 1:
Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node.

Example 2:
Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.

Example 3:
Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteMiddle = function(head) {
  let fast = head;
    let slow = head;
    let dummy = new ListNode(); //to deal with edge case that we need to delete the head node
    dummy.next = head;

    let prev = dummy;

    //both pointers travel
    while(fast && fast.next){
        fast = fast.next.next;
        prev = slow;
        slow = slow.next;
    }

    //delete the node
    while(slow.next){
        slow.val = slow.next.val;
        prev = slow;
        slow = slow.next;
    }

    prev.next = null;
    return dummy.next;
};
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