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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.
刚开始思考这个问题时绕了点弯路，其实很简单，只要把链表中比x小的节点拆下来按顺序链接到一起构成list1，然后把剩下的节点按顺序链接到一起构成list2。最后list1.next=list2即可。代码如下：
1 public class Solution { 2 public ListNode partition(ListNode head, int x) { 3 if(head==null || head.next==null) return head; 4 5 ListNode Partition = head; 6 7 ListNode nhead = new ListNode(0); 8 nhead.next = null; 9 ListNode ntail = nhead; 10 11 ListNode nhead2 = new ListNode(0); 12 nhead2.next = null; 13 ListNode ntail2 = nhead2; 14 15 ListNode p = null; 16 while(Partition!=null){ 17 if(Partition.val < x){ 18 p = Partition.next; 19 ntail.next = Partition; 20 ntail = ntail.next; 21 ntail.next = null; 22 Partition = p; 23 } 24 else{ 25 p = Partition.next; 26 ntail2.next = Partition; 27 ntail2 = ntail2.next; 28 ntail2.next = null; 29 Partition = p; 30 } 31 } 32 ntail.next = nhead2.next; 33 return nhead.next; 34 } 35 }
LeetCode OJ 86. Partition List
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原文地址：http://www.cnblogs.com/liujinhong/p/5404924.html
