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题目：
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：
只要把比x小的节点按顺序连成一条链，比x大或等于的节点连成另一条链，然后把两条链连起来。注意一下边界情况(某条链为空)。
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @param {number} x * @return {ListNode} */ var partition = function(head, x) { if(head==null||head.next==null){ return head; } var lHead=null; var gHead=null; var p=head,pl=null,pg=null,temp=null; while(p){ if(p.val<x){ if(lHead==null){ lHead=p; pl=p; }else{ pl.next=p; pl=pl.next; } }else{ if(gHead==null){ gHead=p; pg=p; }else{ pg.next=p; pg=pg.next; } } temp=p; p=p.next; temp.next=null; } if(pg!=null){ pg.next=null; } if(lHead!=null){ pl.next=gHead; return lHead; }else{ return gHead; } };
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原文地址：http://www.cnblogs.com/shytong/p/5144350.html
