来源
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-linked-list/
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?
题解
迭代
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/class Solution {public ListNode reverseList(ListNode head) {ListNode curr = head, prev = null;while (curr != null) {ListNode temp = curr.next;curr.next = prev;prev = curr;curr = temp;}return prev;}}
复杂度分析
- 时间复杂度:
- 空间复杂度:
递归
递归的经典算法,从后往前反转
/*** 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 || head.next == null) {return head;}ListNode last = reverseList(head.next);head.next.next = head;head.next = null; // 切记此处return last;}}
复杂度分析
- 时间复杂度:
- 空间复杂度:
