LeetCode 206 Reverse Linked List

Problem:

Reverse a singly linked list.

A linked list can be reversed either iteratively or recursively. Could you implement both?

Summary:

翻转链表。

Analysis:

1. Iterative solution

 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* reverseList(ListNode* head) {
12         ListNode *pre = NULL;
13         
14         while (head != NULL) {
15             ListNode *tmp = head->next;
16             head->next = pre;
17             pre = head;
18             head = tmp;
19         }
20         
21         return pre;
22     }
23 };

 2. Recursive solution

若链表为n₁->n₂->...->n_k-1->n_k->n_k+1->...->n_m->NULL

假设n_k+1到n_m的部分已翻转：n₁->n₂->...->n_k-1->n_k->n_k+1<-...<-n_m

我们若要使n_k+1的next指针指向n_k，则我们需要做的是：n_k->next->next = n_k

需要注意的是需要将n₁的next指针指向NULL。

 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* reverseList(ListNode* head) {
12         if (head == NULL || head->next == NULL) {
13             return head;
14         }
15         
16         ListNode *tmp = reverseList(head->next);
17         head->next->next = head;
18         head->next = NULL;
19         
20         return tmp;
21     }
22 };

 

Reference: https://leetcode.com/articles/reverse-linked-list/#approach-1-iterative-accepted

转载于:https://www.cnblogs.com/VickyWang/p/6012874.html