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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.
思路一：
遍历原链表，将比x小的元素拿出来依次插入新的链表，最后将原链表剩下的节点插在新链表后面即得到结果。
1 public ListNode partition(ListNode head, int x) { 2 //dummy节点用来记录原列表的头 3 ListNode dummy = new ListNode(0); 4 dummy.next = head; 5 ListNode pre = dummy; 6 //ret依次存储比x小的节点 7 ListNode ret = new ListNode(0); 8 //tail节点用来记录ret链表的尾部 9 ListNode tail = ret; 10 //遍历原链表 11 while (head != null) { 12 //当前节点值比x小时，插入在tail节点后 13 if (head.val < x) { 14 tail.next = head; 15 tail = tail.next; 16 pre.next = head.next; 17 } else { 18 pre = pre.next; 19 } 20 head = head.next; 21 } 22 //将原链表剩下的节点直接加在tail节点后面 23 tail.next = dummy.next; 24 return ret.next; 25 }
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原文地址：http://www.cnblogs.com/ivanyangzhi/p/4376534.html
