图解精选 TOP 面试题 005.1 | 反转链表之递归求解

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

# 当前节点的下一个节点
next_node = head.next
# 改变下一个节点的指针
next_node.next = head

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head

        # 不断往后获取节点
        cur = self.reverseList(head.next)
        # 反转节点指向
        next_node = head.next
        next_node.next = head
        # 切断当前节点的 next 指向
        head.next = None

        return cur

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }

    cur := reverseList(head.Next)
    nextNode := head.Next
    nextNode.Next = head
    head.Next = nil

    return cur
}