[LeetCode] 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].java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        LinkedList<ListNode> front = new LinkedList<ListNode>();
        LinkedList<ListNode> back = new LinkedList<ListNode>();
        while(head != null){
            ListNode run = head;
            if(run.val < x) front.push(run);
            else back.push(run);
            head = head.next;
        }
        ListNode end = null;
        while(!back.isEmpty()){
            ListNode run = back.pop();
            run.next = end;
            end = run;
        }
        while(!front.isEmpty()){
            ListNode run = front.pop();
            run.next = end;
            end = run;
        }
        return end;
    }
}