题解 | #反转链表#

pre 存储head前一次的值，next存储当前的head.next的值

public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode pre = null;
        ListNode next = null;
        while(head != null){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}