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题目：
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes‘ values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
思路：
Reorder List, reverse linkedList I & II, Sort List,还有好几道leetcode的关于链表的题，其实具体考察的就是三个操作。
LinkedList 的 Partition，Reverse 和 Merge
我当时是认真做了一下LinkedList 的merge sort，就对这些操作比较熟练了。
1 public void reorderList(ListNode head) { 2 ListNode l2 = partition(head); 3 ListNode newL2 = reverse(l2); 4 merge(head, newL2); 5 } 6 7 private ListNode partition (ListNode head) { 8 ListNode left = head; 9 ListNode right = head; 10 ListNode pre = new ListNode(0); 11 while (right != null) { 12 pre = left; 13 left = left.next; 14 right = right.next; 15 if (right != null) { 16 right = right.next; 17 } 18 } 19 pre.next = null; 20 return left; 21 } 22 23 private ListNode reverse(ListNode head) { 24 if (head == null || head.next == null) { 25 return head; 26 } 27 ListNode pre = null; 28 ListNode cur = head; 29 ListNode temp = new ListNode(0); 30 while(cur != null) { 31 temp = cur.next; 32 cur.next = pre; 33 pre = cur; 34 cur = temp; 35 } 36 return pre; 37 } 38 private void merge(ListNode l1, ListNode l2) { 39 ListNode temp = new ListNode(0); 40 while (l1 != null && l2 != null) { 41 temp = l1.next; 42 l1.next = l2; 43 l1 = temp; 44 if (l1 == null) { 45 return; 46 } 47 temp = l2.next; 48 l2.next = l1; 49 l2 = temp; 50 if (l2 == null) { 51 return; 52 } 53 } 54 }
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原文地址：http://www.cnblogs.com/gonuts/p/4518711.html
