[LeetCode Road] Reverse Linked List - Solution/C++

206. Reverse Linked List

Question:

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

Example:

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

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

Input: head = []
Output: []

Constraints:

The number of nodes in the list is the range [0, 5000].

-5000 <= Node.val <= 5000

Source code

Version 1

Idea:
A demonstration is shonw as below:

There are four steps to do: 1) Declare a curr variable that it's a pointer of ListNode to point head ListNode. Shift one position every loop. 2) Change a head->next to prev ListNode. It means proceeding the redirection or revering. Notice: prev ListNode is a new ListNOde with no value (only NULL). 3) Update prev to head. 4) Update head to curr. Repeat (1) - (4) steps until head is NULL.

/**
 * 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* head) {
        if(!head) return head;
        
        ListNode* curr = head;
        ListNode* prev = NULL;
        while(head){
            curr = head->next;
            head->next = prev;
            prev = head;
            head = curr;
        }
        return prev;
    }
};