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.
Example:
Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
题目大意：
将LinkedList利用partition划分为两个部分，LinkedList的前半部分都是比给定的x要小的值，而后半部分是比给定值要大的值。
解法：
将该链表一分为二，分成两个链表，最后再将两个链表串起来。链表一存放的都是比value值要小的节点，而链表二存放的都是比value值要大的节点。
java:
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode h1=new ListNode(0);
ListNode h2=new ListNode(0);
ListNode head1=h1,head2=h2;
while(head!=null){
if(head.val<x){
head1.next=head;
head1=head1.next;
}else{
head2.next=head;
head2=head2.next;
}
head=head.next;
}
head2.next=null;
head1.next=h2.next;
return h1.next;
}
}
原文：https://www.cnblogs.com/xiaobaituyun/p/10676618.html