• 周五. 10月 7th, 2022

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LeetCode题解之Reorder List

admin

11月 28, 2021

1、题目描述

2、题目分析

首先将链表分为两段,然后将后面的一段反转,再合并两个链表。

3、代码

 1 void reorderList(ListNode* head) {
 2         if (head == nullptr || head->next == nullptr || head->next->next == nullptr)
 3             return ;
 4         
 5         ListNode *newH = splitList(head);
 6         ListNode *rp = reverseList(newH);
 7         mergeList(head, rp);   
 8     }
 9     
10     
11    ListNode *reverseList(ListNode* list)
12     {
13         if (list == NULL || list->next == NULL) 
14             return list;
15 
16         ListNode *p = list;
17         ListNode *pn = list->next;
18         list->next = NULL;
19 
20         while (pn != NULL) {
21             ListNode *tmp = pn->next;
22             pn->next = p;
23             p = pn;
24             pn = tmp;
25         } 
26         return p;
27     }
28 
29     void mergeList(ListNode *head_1, ListNode *head_2)
30     {
31         ListNode *p;
32         ListNode *pm = head_1;
33         ListNode *pn = head_2;
34         while (pm != NULL && pn != NULL) {
35             ListNode *tmpm = pm->next;
36             ListNode *tmpn = pn->next;
37             pm->next = pn;
38             if (tmpm != NULL)
39                 pn->next = tmpm;
40             pm = tmpm;
41             pn = tmpn;
42         }
43     }
44 
45     ListNode *splitList(ListNode *head)
46     {
47         if (head == NULL || head->next == NULL)
48             return head;
49         ListNode *pm = head;
50         ListNode *pn = head;
51 
52         while (pn != NULL && pn->next != NULL) {
53             pm = pm->next;
54             pn = pn->next->next;
55 
56         }
57 
58         ListNode *newH = NULL;
59         if (pn == NULL) {
60             newH = pm;
61             ListNode *pt = head;
62             while (pt->next != newH) {
63                 pt = pt->next;
64             }
65             pt->next = NULL;
66         } else if (pn->next == NULL) {
67             newH = pm->next;
68             pm->next = NULL;
69         }
70 
71 
72         return newH;
73     }

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