LeetCode K个一组翻转链表

题目链接：https://leetcode-cn.com/problems/reverse-nodes-in-k-group/

题目大意

　　略。

分析

　　逆转每一段，然后和上一段与下一段衔接即可，加头结点会比较方便。

代码如下

 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* reverseKGroup(ListNode* head, int k) {
12         if(head == NULL || k == 1) return head;
13         
14         int len = getLen(head);
15         ListNode *start = head, *lastTail = new ListNode(0), *newhead = lastTail;
16         newhead->next = head;
17         
18         for(int i = 0; i < len / k; ++i) {
19             lastTail->next = reverseGroup(start, k); // 和上一段尾巴接上
20             lastTail = start;
21             start = lastTail->next;
22         }
23         
24         return newhead->next;
25     }
26     
27     // 返回反转链表的头结点,顺便把尾部接到下一段头部
28     ListNode* reverseGroup(ListNode* head, int k) {
29         ListNode *p1 = head, *p2 = head->next;
30         
31         for(int i = 0; i < k - 1; ++i) {
32             head->next = p2->next;
33             p2->next = p1;
34             p1 = p2;
35             p2 = head->next;
36         }
37         
38         return p1;
39     }
40     
41     int getLen(ListNode *head, ListNode *end = NULL) {
42         int ret = 0;
43         while(head != end) {
44             ++ret;
45             head = head->next;
46         }
47         return ret;
48     }
49 };

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转载于:https://www.cnblogs.com/zaq19970105/p/11351028.html