Sword refers to Offer24-reverse linked list-easy

Question link

Title description:

Define a function, input the head node of a linked list, reverse the linked list and output the head node of the reversed linked list.

For example, enter:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

data range:

0 <= 节点个数 <= 5000

Problem-solving ideas:

In fact, it is a process of turning around
1->2->3->4->5->NULL
NULL <- 1 <- 2 <-3 <- 4 <- 5
1. Set a node1 variable, the initial value is NULL , The initial value of node2 is head
node2 next points to node1

2. Node1 moves to the location of node2, node2 moves to the location of node2next, repeat steps 1

3. Until node2 is NULL, return to node1. The final result is as follows

AC code (c++)

/**
 * 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){
    
    
            return head;
        }
        ListNode * node1 = NULL;
        ListNode * node2 = head;
        while(node2){
    
    
            ListNode * nt = node2->next;
            node2->next = node1;
            node1 = node2;
            node2 = nt;
        }

        return node1;
    }
};