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.
给定一个链表和一个整数n,要求把小于n的放到链表的前面,大于等于n的放到链表的后面,不要打乱原有的顺序。我们可以借助两个辅助节点,smallNode和bigNode分别将链表分为值小于n得一段和值大于等于n的一段,最后将两段连接起来。代码如下:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode partition(ListNode head, int x) { if(head == null) return head; ListNode smallNode = new ListNode(0); ListNode helperS = smallNode; ListNode bigNode = new ListNode(0); ListNode helperB = bigNode; while(head != null) { if(head.val < x) { smallNode.next = head; smallNode = smallNode.next; } else { bigNode.next = head; bigNode = bigNode.next; } head = head.next; } bigNode.next = null; smallNode.next = helperB.next; return helperS.next; } }