Leetcode------Partition List

Title:	Partition List
Pass Rate:	27.5%
Difficulty:	Medium

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 .

The subject is to use a target to divide the list into two parts, less than target and a target greater than

Record with two linked lists, small and big make sure to place the. Next null, otherwise the memory will always overlap. The process of connecting is a pointer redirection process, so be sure to set NULL, or else the entire linked list is connected, the code is as follows:

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {7 * val = x;8 * next = null;9  *     }Ten  * } One  */ A  Public classSolution { -      PublicListNode partition (ListNode head,intx) { -ListNode small=NewListNode (1); theListNode big=NewListNode (1); -ListNode s=small,b=Big; -         if(head==NULL)returnhead; -          while(head!=NULL){ +             if(head.val<x) { -s.next=head; +Head=Head.next; As=S.next; ats.next=NULL; -             } -             Else{ -b.next=head; -Head=Head.next; -b=B.next; inb.next=NULL; -             } to         } +s.next=Big.next; -         returnSmall.next; the     } *}

Leetcode------Partition List