Partition List

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.

遍历原链表，用左右子链表，分别对应大于x小于x，最后把两部分连接起来。

public class Solution {
    //Time: O(n)  Space: O(1)
    public ListNode partition(ListNode head, int x) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode dummyleft = new ListNode(0);
        ListNode dummyright = new ListNode(0);
        ListNode left = dummyleft;
        ListNode right = dummyright;
        
        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = left.next;
            } else {
                right.next = head;
                right = right.next;
            }
            head = head.next;
        }
        
        left.next = dummyright.next;
        right.next = null;
        return dummyleft.next;
    }
}