1. 206题

对有序链表进行反转，如1->2->3，结果为3->2->1

1.1 方法一

初始化pre为None，cur为head，对于1->2进行反转，就是将1的next指向pre，同理，2->3也是，代码如下

# python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre = None
        cur = head
        while cur:
            nex = cur.next
            cur.next = pre
            pre = cur
            cur = nex

        return pre

// java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode cur = head;
        while (cur != null){
            ListNode nex = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nex;
        }
        return pre;
    }
    
}

1.2 方法二：递归

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        rhead = reverseList(head.next)
        head.next.next = head
        head.next = None
        return rhead

// java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        // 递归终止条件
        if(head == null|| head.next == null)
            return head;

        ListNode rhead = reverseList(head.next);

        // head->next此刻指向head后面的链表的尾节点
        // head->next->next = head把head节点放在了尾部
        head.next.next = head;
        head.next = null;

        return rhead;
    }
    
}