206 Reverse Linked List 反转链表
Description:
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?
题目描述:
反转一个单链表。
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？
思路:
- 将链表的结点全部压入栈然后弹出
时间复杂度O(n), 空间复杂度O(n) - 记录结点和结点的后一个结点, 将后一个结点的指针指向该结点
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
ListNode* reverseList(ListNode* head)
{
if (!head or !head -> next) return head;
ListNode* result = NULL;
while (head)
{
ListNode* next = head -> next;
head -> next = result;
result = head;
head = next;
}
return result;
}
};
Java:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if (head == null) return null;
Stack<ListNode> stack = new Stack<>();
while (head.next != null) {
stack.push(head);
head = head.next;
}
ListNode result = head;
while (!stack.empty()) {
result.next = stack.pop();
result = result.next;
}
result.next = null;
return head;
}
}
Python:
# 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:
def reverseListNode(head: ListNode, pre: ListNode) -> ListNode:
if not head:
return pre
next, head.next = head.next, pre
return reverseListNode(next, head)
return reverseListNode(head, None)
