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->...->nk-1->nk->nk+1->...->nm->NULL

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

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

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

posted @ 2016-10-30 23:07  SillyVicky  阅读(188)  评论(0)  编辑  收藏  举报