# Partition List

### 描述​

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.

### 代码​

// Partition List// 时间复杂度O(n)，空间复杂度O(1)class Solution {    public ListNode partition(ListNode head, int x) {        ListNode left_dummy = new ListNode(-1); // 头结点        ListNode right_dummy = new ListNode(-1); // 头结点        ListNode left_cur = left_dummy;        ListNode right_cur = right_dummy;        for (ListNode cur = head; cur != null; cur = cur.next) {            if (cur.val < x) {                left_cur.next = cur;                left_cur = cur;            } else {                right_cur.next = cur;                right_cur = cur;            }        }        left_cur.next = right_dummy.next;        right_cur.next = null;        return left_dummy.next;    }};