[Leedcode 86] Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greate R than or equal to x.

You should preserve the original relative order of the nodes in each of the.

For example,
Given 1->4->3->2->5->2 and x = 3,
Return 1->2->2->4->3->5 .

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode partition (ListNode head,intx) {//The first solution is very simple, the establishment of two two tables, a representative of less than X, 21 represents greater than or equal to X. //iterate through the list, add the tail of the minlist if the node value is less than X, and update the Minlist//It is important to note that the tail node of the maxlist must be empty to prevent the formation of a ring linked list        if(head==NULL)returnHead; /*ListNode minlist=new ListNode (-1);        ListNode minhead=minlist;        ListNode maxlist=new ListNode (-1);        ListNode maxhead=maxlist;                while (Head!=null) {if (head.val>=x) {maxlist.next=head;            Maxlist=head;                }else{Minlist.next=head;            Minlist=head;                    } Head=head.next;        } maxlist.next=null;        Minlist.next=maxhead.next; return minhead.next;*/                /*This solution does not meet the requirements, will destroy the original order ListNode I=head;        ListNode J=head;                while (J!=null) {if (j.val<x) {swap (I,J);                            I=i.next;                    } J=j.next; } return head;*/        //The third solution of the subject is in situ transformation. //First you need to find the first node that is greater than or equal to x and assign it to P2, while P1 points to its predecessor node//then traverse from P2 to the tail of the list, and when the value of P2 next node is less than x, you need to insert the next node of P2 into//P1 's next//The P1 maintains a tail node less than x, and the drawing is clearerListNode Newhead=NewListNode (-1); Newhead.next=Head; ListNode P1=Newhead; ListNode P2=Head; ListNode P1_rear=Head;  while(p2!=NULL&AMP;&AMP;P2.VAL&LT;X) {///////p1=P1.next; P2=P2.next; }         while(p2!=NULL&&p2.next!=NULL){            if(p2.next.val<x) {P1_rear=P1.next; P1.next=P2.next; P2.next=P2.next.next; P1.next.next=p1_rear; P1=P1.next; }Else{P2=P2.next; }        }        returnNewhead.next; }    /*Public void Swap (ListNode I,listnode j) {int temp=i.val;        I.val=j.val;    J.val=temp; }*/    }

[Leedcode 86] Partition List