剑指 Offer 24. 反转链表

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。
解法一：自己想的，用数组存储每个节点，然后再依次next等于前一个

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode[] head2 = new ListNode[5000];;
        int i = 0;
        while (head != null)
        {
            head2[i] = head;
            head = head.next;
            i++;
        }
        if (i > 0)
        {
            head = head2[i-1];
            for (int j = i;j > 0;j--)
            {
                if (j == 1)
                {
                    head2[0].next=null;
                }
                else
                    head2[j-1].next = head2[j-2];
            }
        } 
        return head;

    }
}

解法二：评论区大佬的，迭代法

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null, cur = head, next = null;
        while(cur != null) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}