[sword finger offer simple] sword finger offer 24 Reverse linked list

subject
Code
Execution time ：36 ms, In all Python3 Defeated in submission 72.30% Users of
Memory consumption ：16.1 MB, In all Python3 Defeated in submission 34.25% Users of
Pass the test case ：27 / 27

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        stack=[]
        while head:
            stack.append(head)
            head=head.next
        if not stack:
            return None
        ans=stack[-1]
        now=ans
        stack.pop()
        while stack:
            temp=stack.pop()
            now.next=temp
            now=temp
        now.next=None
        return ans   

【 Method 2： In situ , No extra space 】
Execution time ：40 ms, In all Python3 Defeated in submission 46.78% Users of
Memory consumption ：16.1 MB, In all Python3 Defeated in submission 36.94% Users of
Pass the test case ：27 / 27

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None:
            return None
        pre,now=head,head
        while now:
            if now==pre:
                now=now.next
            else:
                temp_now=now
                temp_now_next=now.next
                now.next=pre
                pre=temp_now
                now=temp_now_next
        head.next=None
        return pre

【 Method 3： Recursive method 】
Execution time ：52 ms, In all Python3 Defeated in submission 9.41% Users of
Memory consumption ：20.6 MB, In all Python3 Defeated in submission 10.97% Users of
Pass the test case ：27 / 27

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

【 Method 4： recursive 】

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        def recur(cur, pre):
            if not cur: return pre     #  Termination conditions 
            res = recur(cur.next, cur) #  Recursive successor node 
            cur.next = pre             #  Modify the node reference to 
            return res                 #  Returns the header node of the inverted linked list 
        
        return recur(head, None)       #  Call recursion and return