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 1:
输入:head = [1,2,3,4,5] 输出:[5,4,3,2,1] 示例 2:
输入:head = [1,2] 输出:[2,1] 示例 3:
输入:head = [] 输出:[]
提示:
链表中节点的数目范围是 [0, 5000] -5000 <= Node.val <= 5000
进阶:链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?
个人题解:
public static ListNode reverseList(ListNode head) {
// 跳出循环标记,当没有下一节点的时候下次跳出循环
boolean pass = false;
// 当前节点
ListNode curr = null;
while (!pass) {
ListNode next = head.next;
if (next == null) {
pass = true;
}
if (curr == null) {
// 初始化第一个尾节点
curr = new ListNode(head.val);
} else {
curr = new ListNode(head.val, curr);
}
head = head.next;
}
return curr;
}
public class ListNode {
public int val;
public ListNode next;
public ListNode() {
}
public ListNode(int val) {
this.val = val;
}
public ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}