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}.
代码: oj 测试通过 248 ms
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 # @param head, a ListNode 9 # @return nothing 10 def reorderList(self, head): 11 if head is None or head.next is None or head.next.next is None: 12 return head 13 14 dummyhead = ListNode(0) 15 dummyhead.next = head 16 17 # get the length of the linked list 18 p = head 19 list_length = 0 20 while p is not None: 21 list_length += 1 22 p = p.next 23 24 #reverse the second half linked list 25 fast = dummyhead 26 for i in range((list_length+1)/2): 27 fast = fast.next 28 pre = fast 29 curr = pre.next 30 for i in range( (list_length)/2 - 1 ): 31 tmp = curr.next 32 curr.next = tmp.next 33 tmp.next = pre.next 34 pre.next = tmp 35 36 #merge 37 h2 = pre.next 38 fast.next = None # cut the connection between 1st half linked list and 2nd half linked list 39 while head is not None and h2 is not None: 40 tmp = head.next 41 head.next = h2 42 tmp2 = h2.next 43 head.next.next = tmp 44 h2 = tmp2 45 head = tmp 46 47 return dummyhead.next
思路：
这道题的路子分三块：
1. 遍历单链表 求链表长度
2. 锁定后半个链表，反转后半个链表的每个元素
3. 切断前后半个链表的链接处 然后合并两个链表