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.

public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode safe = new ListNode(x-1);
        ListNode pre = head;
        ListNode cur = safe;
        safe.next = head;
        while(pre!=null && pre.val<x){
            cur = pre;
            pre = pre.next;
        }
        if(pre==null){
            return head;
        }
        else{
            ListNode p = pre;
            while(pre!=null){
                if(pre.val<x){
                    ListNode temp = cur.next;
                    p.next =pre.next;
                    cur.next = pre;
                    pre.next = temp;
                    cur = pre;
                    pre =p.next;
                }
                else{
                    p = pre;
                    pre = pre.next;
                }
            }
        }
        return safe.next;
    }
}

 

ListNode post =null;
if(cur!=null){
post = cur.next;
}