[leetcode] 206. Reverse Linked List @ python

原题

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?

解法1

遍历法, 先遍历链表, 将链表转化为list, 然后反转list, 将list转化为链表.
Time: 2*O(n)
Space: O(n)

代码

class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        l = self.getList(head)
        l.reverse()
        dummy = p = ListNode(0)
        for val in l:
            p.next = ListNode(val)
            p = p.next
            
        return dummy.next
        
        
    def getList(self, head):
        l = []
        while head:
            l.append(head.val)
            head = head.next
        return l

解法2

递归. Base case是当head是空或者是单独节点时, 直接返回原head. 当我们移动到最后一个Node以后, 将这个Node定义为我们新的head, 取名new_head. 我们从new_head开始重复性的往前更改指针的方向.
Time: O(n)
Space: O(1)

代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # base case
        if not head or not head.next:
            return head
        
        new_head = self.reverseList(head.next)
        next_node = head.next
        next_node.next = head
        head.next = None
        return new_head

解法3

迭代法

代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: 'ListNode') -> 'ListNode':
        prev = None
        while head:
            curr = head
            head = head.next
            curr.next = prev
            prev = curr
            
        return prev