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本专栏无以上特殊说明的其他Sol是自己写的

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

Iterative Sol

class Solution:
    def reverseList(self, head):
        prev = None
        curr = head
        while curr != None:
            nextTemp = curr.next
            curr.next = prev
            prev = curr
            curr = nextTemp
        return prev

Recursive Sol

class Solution:
    def reverseList(self, head):
        if head is None or head.next is None:
            return head
        p = self.reverseList(head.next)
        head.next.next = head;
        head.next = None
        return p