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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.
思路：把list分成两份，一份是比pivot小，一份是比pivot大，然后连接起来。用dummy node来partition list，因为head大小无法判断。注意最后要把快的list尾巴断开，用null来断，不然可能还连接着其他node。
/** * 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||head.next==null)return head; ListNode dummy1=new ListNode(-1); ListNode dummy2=new ListNode(-1); ListNode copydummy1=dummy1; ListNode copydummy2=dummy2; while(head!=null){ if(head.val<x){ copydummy1.next=head; copydummy1=copydummy1.next; }else{ copydummy2.next=head; copydummy2=copydummy2.next; } head=head.next; } copydummy2.next=null; copydummy1.next=dummy2.next; return dummy1.next; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5953215.html
