Partition List

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.

For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.

Solution

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
          if (head == null) return null;

        ListNode dummy1 = new ListNode(1);
        ListNode dummy2 = new ListNode(1);

        ListNode p = dummy1;
        ListNode q = dummy2;

        ListNode runner = head;
        while (runner != null) {
            if (runner.val < x) {
                p.next = runner;
                p = p.next;
            } else {
                q.next = runner;
                q = q.next;
            }

            runner = runner.next;
        }

        q.next = null;
        p.next = dummy2.next;
        return dummy1.next;
    }
}