Reverse Linked List

Tags : linkedlist, leetcode, cpp, easy

Given the head of a singly linked list, reverse the list, and return the reversed list.

Examples #

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

Constraints #

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

Solutions #

#1 #


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* cn, ListNode* pn){
        if(!cn) return pn;
        ListNode* temp = reverseList(cn->next,cn);
        cn->next = pn;
        return temp;
    }
    ListNode* reverseList(ListNode* head) {
        return reverseList(head,NULL);
    }
};

#2 #


class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* curr = head;
        ListNode* prev = nullptr;
        while (curr != nullptr) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
};