Leetcode 86. Partition List

题目：
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.

思路：创建一个存储大结点的链表的伪头，创建一个存储小结点的链表的伪头，从给定链表的头结点开始遍历，小于给定的数就把该结点附加到小结点链表的尾部，否则附加到大结点链表的尾部，最后把大结点的链接到小结点链表的尾部。

具体实现：

/** * 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 max = new ListNode(0);
        ListNode min = new ListNode(0);
        ListNode pmin = min;
        ListNode pmax = max;
        ListNode p = head;
        while(p!=null){
            ListNode temp = p;
            p = p.next;
            temp.next = null;
            if(temp.val < x){
                pmin.next = temp;
                pmin = pmin.next;
            }else{
                pmax.next = temp;
                pmax = pmax.next;
            }
        }
        pmin.next = max.next;
        return min.next;
    }
}