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.

给定一个链表和一个整数n，要求把小于n的放到链表的前面，大于等于n的放到链表的后面，不要打乱原有的顺序。我们可以借助两个辅助节点，smallNode和bigNode分别将链表分为值小于n得一段和值大于等于n的一段，最后将两段连接起来。代码如下：

/**
 * 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 head;
        ListNode smallNode = new ListNode(0);
        ListNode helperS = smallNode;
        ListNode bigNode = new ListNode(0);
        ListNode helperB = bigNode;
        while(head != null) {
            if(head.val < x) {
                smallNode.next = head;
                smallNode = smallNode.next;
            } else {
                bigNode.next = head;
                bigNode = bigNode.next;
            }
            head = head.next;
        }
        bigNode.next = null;
        smallNode.next = helperB.next;
        return helperS.next;
    }
}