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
.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode partition(ListNode head, int x) { ListNode lin = head; ListNode small = new ListNode(0); ListNode hs = small; ListNode large = new ListNode(0); ListNode hl = large; while(lin!=null){ ListNode ln = new ListNode(lin.val); if(lin.val<x){ small.next = ln; small = small.next; }else{ large.next = ln; large = large.next; } lin = lin.next; } small.next = hl.next; return hs.next; } }
原文地址:http://blog.csdn.net/guorudi/article/details/41120505