leetcode 206 Reserve Linked List(反转链表) python3 递归&迭代思路

所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。

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

class Solution:
    def reverseList(self, head, prev = None):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        
        # 迭代思路：重复一定的算法，达到想要的目的。数学上二分法，牛顿法是很好的迭代例子
        # while head:
        #     curr = head
        #     head = head.next
        #     curr.next = prev
        #     prev = curr
        # return prev
        
        
        # 递归思路：在return处调用自己（尾递归）
        if not head:
            return prev
        
        curr, head.next = head.next, prev    # 新旧链表的两个方向同时前进
        return self.reverseList(curr, head)
            
        
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。