## 一、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`.

## 三、解题代码

``````/**
* 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) {
ListNode leftDummy = new ListNode(0);
ListNode leftCurr = leftDummy;
ListNode rightDummy = new ListNode(0);
ListNode rightCurr = rightDummy;

ListNode runner = head;
while (runner != null) {
if (runner.val < x) {
leftCurr.next = runner;
leftCurr = leftCurr.next;
} else {
rightCurr.next = runner;
rightCurr = rightCurr.next;
}
runner = runner.next;
}

// cut off ListNode after rightCurr to avoid cylic
rightCurr.next = null;
leftCurr.next = rightDummy.next;

return leftDummy.next;
}
}``````