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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *pre = NULL;
        
        while (head != NULL) {
            ListNode *tmp = head->next;
            head->next = pre;
            pre = head;
            head = tmp;
        }
        
        return pre;
    }
};

 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。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        
        ListNode *tmp = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        
        return tmp;
    }
};

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