LeetCode-206-反转链表
题目
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?
解法
迭代解法:
public ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode current = head;
while(current != null) {
ListNode next = current.next;
current.next = pre;
pre = current;
current = next;
}
return pre;
}
运行结果:
递归解法
- head.next 之下的节点都已经反转好
- head.next.next = head 节点
- head.next 置为null
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode p = reverseList(head.next);
head.next.next = head;
head.next = null;
return p;
}
运行结果:
标题:LeetCode-206-反转链表
作者:guobing
地址:http://www.guobingwei.tech/articles/2020/11/07/1604679975730.html