假设我们有一个链表,我们必须扭转它。因此,如果列表类似于1→3→5→7,则新的反向列表将为7→5→3→1
为了解决这个问题,我们将遵循这种方法-
定义一个过程以递归的方式执行列表逆转,以解决(头,尾)
如果头部不存在,则返回头部
temp:= head.next
head.next:=返回
后=头
如果温度为空,则返回头
头=温度
返回解决(头,后)
让我们看下面的实现以更好地理解-
class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
head = "ListNode(elements[0])
for element in elements[1:]:
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
return head
def print_list(head):
ptr = head
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
return self.solve(head,None)
def solve(self, head, back):
if not head:
return head
temp= head.next
#print(head.val)
head.next = back
back = head
if not temp:
return head
head = temp
return self.solve(head,back)
list1 = make_list([1,3,5,7])
ob1 = Solution()list2 = ob1.reverseList(list1)
print_list(list2)
list1 = [1,3,5,7]
输出结果
[7, 5, 3, 1, ]