LeetCode: Reverse Linked List II

最新推荐文章于 2025-09-13 19:30:09 发布

weixin_30596023

最新推荐文章于 2025-09-13 19:30:09 发布

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CC 4.0 BY-SA版权

文章标签： c/c++ c#

原文链接：http://www.cnblogs.com/yingzhongwen/archive/2013/04/20/3031946.html

本文提供了一种反转链表中指定区间[m, n]的方法。通过迭代的方式调整节点指针，实现链表部分区间的反转。适用于C++及C#环境。

没做出来，看网上答案，这题难度在于编程

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *reverseBetween(ListNode *head, int m, int n) {
12         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         if (!head) return NULL;
15         ListNode *q = NULL;
16         ListNode *p = head;
17         for (int i = 0; i < m-1; i++) {
18             q = p;
19             p = p->next;
20         }
21         ListNode *end = p;
22         ListNode *pPre = p;
23         p = p->next;
24         for (int i = m+1; i <= n; i++) {
25             ListNode *pNext = p->next;
26             p->next = pPre;
27             pPre = p;
28             p = pNext;
29         }
30         end->next = p;
31         if (q) q->next = pPre;
32         else head = pPre;
33         return head;
34     }
35 };

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public int val;
 5  *     public ListNode next;
 6  *     public ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode ReverseBetween(ListNode head, int m, int n) {
11         if (head == null) return null;
12         ListNode q = null, p = head;
13         for (int i = 0; i < m-1; i++) {
14             q = p;
15             p = p.next;
16         }
17         ListNode end = p, pPre = p;
18         p = p.next;
19         for (int i = m+1; i <= n; i++) {
20             ListNode pNext = p.next;
21             p.next = pPre;
22             pPre = p;
23             p = pNext;
24         }
25         end.next = p;
26         if (q != null) q.next = pPre;
27         else head = pPre;
28         return head;
29     }
30 }