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) { if (head == null) { return null; } ListNode dummyLeft = new ListNode(0); ListNode dummyRight = new ListNode(0); ListNode Left = dummyLeft; ListNode Right = dummyRight; while(head!=null) { if(head.val<x) { Left.next = head; Left = head; //Left = Left.next; } else { Right.next = head; Right = head; //Right = Right.next; } head = head.next; } Right.next=null; Left.next = dummyRight.next; return dummyLeft.next; } }
原文：http://www.cnblogs.com/hygeia/p/4897468.html