Leetcode 206.反转链表

Posted on |

206.反转链表

题目

反转一个单链表。

1
2
3
4
示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

方法

方法1：迭代

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
# 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
        """
        if head is None or head.next is None:
            return head
        last = None
        while head:
            tmp = head.next
            head.next = last
            last = head
            head = tmp
        return last

方法2：递归

假设链表$n_1 \to n_2 \to n_{k-} \to n_k \to n_{k+1} \to …\to n_m \to None$中，结点$n_{k+1}$到$n_m$已经反转了，当前结点就是$n_k$

现在只需要将$n_{k+1}$的下个结点设为$n_k$即可：$n_k.next.next=n_k$
当然，还需要将$n_1.next=None$

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
# 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
        """
        if not head:
            return
        if not head.next:
            return head
        node = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return node