LeetCode OJ 86. 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.

刚开始思考这个问题时绕了点弯路，其实很简单，只要把链表中比x小的节点拆下来按顺序链接到一起构成list1，然后把剩下的节点按顺序链接到一起构成list2。最后list1.next=list2即可。代码如下：

public class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head==null || head.next==null) return head;
        
        ListNode Partition = head;
        
        ListNode nhead = new ListNode(0);
        nhead.next = null;
        ListNode ntail = nhead;
        
        ListNode nhead2 = new ListNode(0);
        nhead2.next = null;
        ListNode ntail2 = nhead2;
        
        ListNode p = null;
        while(Partition!=null){
            if(Partition.val < x){
                p = Partition.next;
                ntail.next = Partition;
                ntail = ntail.next;
                ntail.next = null;
                Partition = p;
            }
            else{
                p = Partition.next;
                ntail2.next = Partition;
                ntail2 = ntail2.next;
                ntail2.next = null;
                Partition = p;
            }
        }
        ntail.next = nhead2.next;
        return nhead.next;
    }
}