难度：Easy
题目要求
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?
中文说明
反转一个单链表。
题解
迭代法
通过维护三个指针，分别指向prev、cur、next，循环遍历每个节点时，对其进行翻转。注意设置初始的prev为null。
public class Solution {
public ListNode reverseList(ListNode head) {
ListNode prev, cur, next;
prev = null;
cur = head;
while (cur != null) {
next = cur.next;
cur.next = prev;
prev = cur;
cur = next;
}
return prev;
}
}
时间复杂度：O(N)
空间复杂度：O(1)
递归法
class Solution {
public ListNode reverseList(ListNode head) {
return helper(head, null);
}
private ListNode helper(ListNode head, ListNode newHead) {
if (head == null) return newHead;
ListNode next = head.next;
head.next = newHead;
return helper(next, head);
}
}
- 时间复杂度：O(N)
- 空间复杂度：O(1)（尾递归）
