Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both? 链接列表可以反复或递归地反转。 你能同时实施吗？
思路： 有点类似于栈的思想 先进后出
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
stack = []
i = 0
p = head
while p:
stack.append(p.val)
#print(p.val)
p = p.next
i += 1
#print(stack)
q = head
l = []
for j in range(len(stack)-1, -1, -1):
#print(j)
#q.val = stack[j]
l.append(stack[j])
#print(q.val)
#q = q.next
#print(l)
return l
if __name__ == "__main__":
L1 = ListNode(1)
L1.next = ListNode(2)
L1.next.next = ListNode(3)
L1.next.next.next = ListNode(4)
L1.next.next.next.next = ListNode(5)
Solution().reverseList(L1)
通过，输出的形式必须是数组，才会通过
网上思路： https://blog.csdn.net/coder_orz/article/details/51306170
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
p = head
newList = []
while p:
newList.insert(0, p.val)
p = p.next
p = head
for v in newList:
p.val = v
print(p.val)
p = p.next
return head
if __name__ == "__main__":
L1 = ListNode(1)
L1.next = ListNode(2)
L1.next.next = ListNode(3)
L1.next.next.next = ListNode(4)
L1.next.next.next.next = ListNode(5)
print(Solution().reverseList(L1))
疑问：为啥是return head
