【LeetCode】206.反转链表
206.反转链表
知识点:链表;双指针;
题目描述
给你单链表的头节点 head ,请你反转链表,并返回反转后的链表。
示例
输入:head = [1,2,3,4,5]
输出:[5,4,3,2,1]
输入:head = [1,2]
输出:[2,1]
输入:head = []
输出:[]
解法一:双指针
将链表反转,就是把箭头的顺序换一下就可以了,可以采用双指针的方法,让cur的next为pre,然后pre和cur逐渐遍历,直到最后;
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode cur = head;
while(cur != null){
ListNode temp = cur.next; //临时存储当前的下一节点,要不断了就找不到了;
cur.next = pre; //往前指
pre = cur; //指针往后移;
cur = temp;
}
return pre;
}
}
- python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
prehead = None
pre = prehead
cur = head
while cur:
next = cur.next
cur.next = pre
pre = cur
cur = next
return pre
时间复杂度:O(N);
体会
链表的题一般都需要定义虚头,并且格外关注前一节点和后一节点。