86. Partition List

Problem

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

Solution

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function(head, x) {
  var l1 = new ListNode(0);
  var l2 = new ListNode(0);
  var now1 = l1;
  var now2 = l2;
  var now = head;

  while (now) {
    if (now.val < x) {
      now1.next = now;
      now1 = now1.next;
    } else {
      now2.next = now;
      now2 = now2.next;
    }
    now = now.next;
  }

  now1.next = l2.next;
  now2.next = null;

  return l1.next;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).