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}
.
比较容易思考且实现的一个思路是, 将链表从中心拆成两半,后一半全部反向连接,然后前一半和后一半一个一个节点交替连接。
拆分阶段使用快慢指针方法即可得到两个平分的链表。
反向和合并阶段就是很常规的链表操作。
代码:
1 public void reorderList(ListNode head) { 2 if(head==null || head.next==null) 3 return; 4 //split 5 ListNode fast = head; 6 ListNode slow = head; 7 while(true) 8 { 9 fast = fast.next.next; 10 if(fast==null || fast.next==null) 11 break; 12 slow = slow.next; 13 } 14 fast = slow.next; 15 slow.next = null; 16 slow = head; 17 //reverse fast 18 fast = reverse(fast); 19 //insert 20 while(slow!=null && fast!=null) 21 { 22 ListNode snext=slow.next; 23 ListNode fnext=fast.next; 24 slow.next=fast; 25 if(snext!=null) 26 fast.next=snext; 27 slow = snext; 28 fast = fnext; 29 } 30 } 31 public ListNode reverse(ListNode head) 32 { 33 ListNode cur = head; 34 ListNode pre = null; 35 while(cur!=null) 36 { 37 ListNode temp = cur.next; 38 cur.next = pre; 39 pre = cur; 40 cur = temp; 41 } 42 return pre; 43 }