leetcode------Partition List

标题：	Partition List
通过率：	27.5%
难度：	中等

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.

本题就是利用一个target将链表分成两部分，小于target和大于target的

用两个链表进行记录，small和big进行连接时候一定要将.next置null，否则内存会一直叠加。连接的过程就是指针的重定向过程，所以一定要置null，要不然会讲整个链表连接过来，代码如下：

/**
 * 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 small=new ListNode(1);
        ListNode big=new ListNode(1);
        ListNode s=small,b=big;
        if(head==null)return head;
        while(head!=null){
            if(head.val<x){
                s.next=head;
                head=head.next;
                s=s.next;
                s.next=null;
            }
            else{
                b.next=head;
                head=head.next;
                b=b.next;
                b.next=null;
            }
        }
        s.next=big.next;
        return small.next;
    }
}