86. 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.

思路：把list分成两份，一份是比pivot小，一份是比pivot大，然后连接起来。用dummy node来partition list，因为head大小无法判断。注意最后要把快的list尾巴断开，用null来断，不然可能还连接着其他node。

/**
 * 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||head.next==null)return head;
    ListNode dummy1=new ListNode(-1);
    ListNode dummy2=new ListNode(-1);
    ListNode copydummy1=dummy1;
    ListNode copydummy2=dummy2;
    while(head!=null){
        if(head.val<x){
            copydummy1.next=head;
            copydummy1=copydummy1.next;
        }else{
            copydummy2.next=head;
            copydummy2=copydummy2.next;
        }
        head=head.next;
    }
    copydummy2.next=null;
    copydummy1.next=dummy2.next;
    return dummy1.next;
    }
}