Partition List

Link: http://oj.leetcode.com/problems/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.

/**
 * 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) {
        ListNode remains = new ListNode(0);
        ListNode remains_point = remains;
        ListNode result = new ListNode(0);
        ListNode result_point = result;
        while(head!=null){
            if(head.val<x) {
                result.next = head;
                result = result.next;
                head = head.next;
            }else {
                remains.next=head;
                remains = remains.next;
                head = head.next;
            }
        }
        if(remains_point.next!=null){
            result.next=remains_point.next;
            //Note we should end the remains link list
            remains.next=null;
        }
        return result_point.next;
    }
}

第一题提交没通过。因为remains链表没有加终止。