LeetCode Notes_#92 Reverse Linked List II（C++,Python）

LeetCode Notes_#92 Reverse Linked List II（C++,Python）

LeetCode 

Contents

• 题目
• 思路和解答
  • 思路
  • 解答
    • C++
    • Python

题目

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

思路和解答

思路

1. 几个关键节点

关键节点

2. 整体思路

图解

图解

（1）先将指针head移动到开始逆置的节点modify_list_tail
（2) 按照之前#206 Reverse Linked List的做法，将[m,n]这一段逆置
（3）将被逆置的一段和逆置段前驱，逆置段后继连接起来
（4）考虑特殊情况：m=1?
如果是从第一个节点开始逆置，那么最后就不需要连接pre_head了

（5）为什么逆置的最后一个节点不需要考虑？

• 因为n<=length,之后无论如何还有一个NULL节点，最后还是要把链表和NULL接起来

3.思考：返回结果是什么？

• 返回的结果的形式是指向链表第一个节点的头指针
• 如果m!=1,结果是一开始的head,要提前保存；如果m=1，结果是new_head

解答

C++

1. class Solution { 

   ---

2. public: 

   ---

3. ListNode* reverseBetween(ListNode* head, int m, int n) { 

   ---

4. int change_len = n-m+1; 

   ---

5. ListNode * new_head = NULL; 

   ---

6. ListNode* result = head; 

   ---

7. ListNode* pre_head = NULL; 

   ---

8. ListNode* modify_list_tail = NULL; 

   ---

9. for(int i=1;i<m;i++){ 

   ---

10. pre_head = head;//保存逆置段的前驱 

    ---

11. head=head->next;//将head移动到m节点  

    ---

12. } 

    ---

13. modify_list_tail = head; 

    ---

14. while(change_len&&head){//将逆置段逆置 

    ---

15. ListNode* next = head->next; 

    ---

16. head->next = new_head; 

    ---

17. new_head = head; 

    ---

18. head = next;//最后一次循环，head就是逆置段的后置 

    ---

19. change_len--; 

    ---

20. } 

    ---

21.  

    ---

22. modify_list_tail->next = head; 

    ---

23. if (pre_head){ 

    ---

24. pre_head->next = new_head; 

    ---

25. } 

    ---

26. else{ 

    ---

27. result = new_head; 

    ---

28. } 

    ---

29. return result; 

    ---

30. } 

    ---

31. }; 

    ---

4ms,C++的确比较快

一个困惑的点：
问题:16行,new_head是NULL,那么是不是最后的结果会在逆置段和逆置后继之间有一个NULL?
回答:不会，因为22行连接逆置段后继的时候，直接将NULL修改为逆置段后继了

自己编写测试用例，在VS Studio里调试

1. #include <iostream> 

   ---

2. using namespace std; 

   ---

3. struct ListNode { 

   ---

4. int val; 

   ---

5. ListNode *next; 

   ---

6. ListNode(int x) : val(x), next(NULL) {} 

   ---

7. }; 

   ---

8. class Solution { 

   ---

9. public: 

   ---

10. ListNode* reverseBetween(ListNode* head, int m, int n) { 

    ---

11. int change_len = n - m + 1; 

    ---

12. ListNode * new_head = NULL; 

    ---

13. ListNode* result = head; 

    ---

14. ListNode* pre_head = NULL; 

    ---

15. ListNode* modify_list_tail = NULL; 

    ---

16. for (int i = 1; i<m; i++) { 

    ---

17. pre_head = head;//保存逆置段的前驱 

    ---

18. head = head->next;//将head移动到m节点  

    ---

19. } 

    ---

20. modify_list_tail = head; 

    ---

21. while (change_len&&head) {//将逆置段逆置 

    ---

22. ListNode* next = head->next; 

    ---

23. head->next = new_head; 

    ---

24. new_head = head; 

    ---

25. head = next;//最后一次循环，head就是逆置段的后置 

    ---

26. change_len--; 

    ---

27. } 

    ---

28.  

    ---

29. modify_list_tail->next = head; 

    ---

30. if (pre_head) { 

    ---

31. pre_head->next = new_head; 

    ---

32. } 

    ---

33. else { 

    ---

34. result = new_head; 

    ---

35. } 

    ---

36. return result; 

    ---

37. } 

    ---

38. }; 

    ---

39.  

    ---

40. void main() { 

    ---

41. ListNode a(1); 

    ---

42. ListNode b(2); 

    ---

43. ListNode c(3);  

    ---

44. ListNode d(4);  

    ---

45. ListNode e(5); 

    ---

46. a.next = &b; 

    ---

47. b.next = &c; 

    ---

48. c.next = &d; 

    ---

49. d.next = &e; 

    ---

50. Solution solve; 

    ---

51. ListNode *head = solve.reverseBetween(&a, 2, 4); 

    ---

52. while (head) { 

    ---

53. printf("%d ", head->val); 

    ---

54. head = head->next; 

    ---

55. } 

    ---

56. system("pause"); 

    ---

57.  

    ---

58. } 

    ---

Python

1. class Solution(object): 

   ---

2. def reverseBetween(self, head, m, n): 

   ---

3. """ 

   ---

4. :type head: ListNode 

   ---

5. :type m: int 

   ---

6. :type n: int 

   ---

7. :rtype: ListNode 

   ---

8. """ 

   ---

9. change_len = n-m+1 

   ---

10. pre_head = None 

    ---

11. result = head 

    ---

12.  

    ---

13. for i in range(1,m): 

    ---

14. pre_head = head 

    ---

15. head = head.next 

    ---

16. modify_list_tail = head 

    ---

17.  

    ---

18. new_head = None 

    ---

19. while(change_len and head): 

    ---

20. next = head.next 

    ---

21. head.next = new_head 

    ---

22. new_head = head 

    ---

23. head = next 

    ---

24. change_len-=1 

    ---

25. # if modify_list_tail: 

    ---

26. modify_list_tail.next = head 

    ---

27.  

    ---

28. if(pre_head): 

    ---

29. pre_head.next = new_head 

    ---

30. else: 

    ---

31. result = new_head 

    ---

32.  

    ---

33. return result 

    ---

32ms