Leetcode#143 Reorder List

原题地址

 

先把链表分割成前后两半，然后交叉融合

实践证明，凡是链表相关的题目，都应该当成工程类题目做，局部变量、功能函数什么的随便整，代码长了没关系，关键是清楚，不容易出错。

 

代码：

 1 ListNode *reverseList(ListNode *head) {

 2   if (!head) return head;

 3         

 4   ListNode *prev = head;

 5   head = head->next;

 6   prev->next = NULL;

 7   while (head) {

 8     ListNode *next = head->next;

 9     head->next = prev;

10     prev = head;

11     head = next;

12   }

13         

14   return prev;

15 }

16     

17 ListNode *mergeList(ListNode *head1, ListNode *head2) {

18   ListNode *next1, *next2;

19   ListNode *head = head1;

20         

21   while (head1 && head2) {

22     next1 = head1->next;

23     next2 = head2->next;

24     head1->next = head2;

25     if (next1)

26       head2->next = next1;

27     head1 = next1;

28     head2 = next2;

29   }

30         

31   return head;

32 }

33 

34 void reorderList(ListNode *head) {

35   if (!head)

36     return;

37             

38   ListNode *fast = head;

39   ListNode *slow = head;

40   while (fast && fast->next) {

41     fast = fast->next->next;

42     slow = slow->next;

43   }

44         

45   ListNode *head1 = head;

46   ListNode *head2 = reverseList(slow->next);

47   slow->next = NULL;

48   mergeList(head1, head2);

49 }