Partition List 分隔链表
Description
Given the head of 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.
将一个链表，按照给定的x值，分成2部分，链表前部小于x，后部大于等于x
输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]
思路: 虚头
新建链表a,b,并利用p遍历链表，将小于x的节点放入a，大于等于x的节点放入b，然后连接ab。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode bhead = new ListNode(0, null);//small
ListNode ahead = new ListNode(0, null);//big
ListNode atail=ahead,btail=bhead;
ListNode p = head;
ListNode q = null;
while(p!=null){
q = p.next;
if (p.val < x) {
p.next = atail.next;
atail.next = p;
atail = p;
}
else{
p.next = btail.next;
btail.next = p;
btail = p;
}
p = q;
}
atail.next = bhead.next;
return ahead.next;
}
}