【LeetCode】手撕系列—206. 反转链表_三个指针

• 原题链接：206. 反转链表

1- 思路

• 定义三个指针
• prev：记录 cur 的前一个结点
• cur：记录当前需要翻转的结点
• tmp：记录 cur 的 next 防止遍历过程中找不到链表的剩余部分

2- 题解

⭐反转链表 ——题解思路

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode cur = head;
        ListNode tmp = null;
        while(cur!=null){
            tmp = cur.next;

            cur.next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
    }
}

3-ACM模式

public class reverseLink {

    static class ListNode{
        int val;
        ListNode next;
        ListNode(){}
        ListNode(int x){
            val = x;
        }
    }

    public static ListNode reverse(ListNode head){
        ListNode prev = null;
        ListNode cur = head;
        ListNode tmp = null;
        while(cur!=null){
            tmp = cur.next;

            cur.next = prev;
            prev = cur;
            cur = tmp;
        }
        return prev;
    }


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("输入链表长度n");
        int n = sc.nextInt();
        ListNode head=null ,tail=null;
        for(int i=0; i < n;i++){
            ListNode newNode = new ListNode(sc.nextInt());
            if(head==null){
                head = newNode;
                tail = newNode;
            }else{
                tail.next = newNode;
                tail = newNode;
            }
        }
        ListNode forRes  = reverse(head);

        System.out.println("翻转后的链表为：");
        while(forRes!=null){
            System.out.print(forRes.val+" ");
            forRes = forRes.next;
        }
    }
}