Leetcode Partition List 分割鏈表

題目：

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.

分析：

對於值大於等於x的節點，直接插到鏈表。對於值小於x的節點，需要找到第一個大於等於x的節點，插到它的前面。

Java代碼實現：

/**
 * 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) {
        if(head==null || head.next==null)
            return head;
            
        ListNode dummy = new ListNode(0);
        ListNode node = dummy;
        node.next = null;

        while(head!=null)
        {
            if(head.val>=x)
            {
                while(node.next!=null)
                {
                    node = node.next;
                }
                node.next = new ListNode(head.val);
                node.next.next = null;
            }
            else
            {
                while(node.next!=null && node.next.val < x)
                {
                    node = node.next;
                }
                ListNode temp = node.next;
                node.next = new ListNode(head.val);
                node.next.next = temp;
            }
            node = dummy;
            head = head.next;
        }
        
        return dummy.next;
    }
}