链表反转实现

方法一：双指针法

采用就地反转，不用额外开辟空间

注意：接收链表的head节点，返回当前节点

方法二：递归法

代码如下：

class Node(object):
    def __init__(self, item, next=None):
        self.item = item
        self.next = next

def create_linklist_head(li):
    head = Node(li[0])
    for e in li[1:]:
        node = Node(e)
        node.next = head
        head = node
    return head

def create_linklist_tail(li):
    head = Node(li[0])
    tail = head
    for e in li[1:]:
        node = Node(e)
        tail.next = node
        tail = node
    return head

def print_linklist(lk):
    while lk:
        print(lk.item, end=',')
        lk = lk.next

def inverse_linklist(head):
    if head == None or head.next == None:
        return head
    currNode = head
    predNode = None
    while currNode:
        tmp = currNode.next
        currNode.next = predNode
        predNode = currNode
        currNode = tmp
    return predNode

def recursion_inverse_lk(head):
    if head == None or head.next == None:
        return head
    new_head = recursion_inverse_lk(head.next)
    head.next.next = head
    head.next = None
    return new_head


if __name__ == '__main__':
    lk = create_linklist_tail([1,2,3,4,5])
    print('原始：')
    print_linklist(lk)
    # res = inverse_linklist(lk)
    res = recursion_inverse_lk(lk)
    print('\n反转：')
    print_linklist(res)

结果展示：