【面试题】反转单链表

面试题 反转单链表
1、迭代法

2、递归法

Java 实现

class ListNode {
    int val;
    ListNode next;

    ListNode(int val) {
        this.val = val;
        next = null;
    }

    @Override
    public String toString() {
        return val + "->" + next;
    }
}

public class ReverseListedList {
    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        ListNode a = new ListNode(2);
        ListNode b = new ListNode(3);
        ListNode c = new ListNode(4);
        ListNode d = new ListNode(5);
        head.next = a;
        a.next = b;
        b.next = c;
        c.next = d;

        System.out.println("链表反转前：" + head);
        head = reverseListByIterative(head);
        System.out.println("迭代反转后：" + head);

        head = reverseListByRecursive(head);
        System.out.println("递归反转后：" + head);
    }

    public static ListNode reverseListByIterative(ListNode head) {
        if (head == null)
            return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy.next;
        ListNode pcur = prev.next;
        while (pcur != null) {
            prev.next = pcur.next;
            pcur.next = dummy.next;
            dummy.next = pcur;
            pcur = prev.next;
        }
        return dummy.next;
    }

    public static ListNode reverseListByRecursive(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = reverseListByRecursive(head.next);
        head.next.next = head;
        head.next = null;
        return newHead;
    }
}

运行结果

链表反转前：1->2->3->4->5->null
迭代反转后：5->4->3->2->1->null
递归反转后：1->2->3->4->5->null