面试题-链表翻转

题目

给一个链表，进行反转

解题思路

每个都插头就行了

public class Main2 {


    static class Node {
        int index;
        Node next;

        public Node() {
        }

        public Node(int index, Node next) {
            this.index = index;
            this.next = next;
        }
    }


    static Node init() {
        Node head = new Node(-1, null);
        Node tail = new Node(-1, null);
        Node temp = head;
        Node cur = head;
        for (int i = 1; i <= 100; ++i) {
            temp = new Node(i, null);
            cur.next = temp;
            cur = cur.next;
        }
        cur.next = tail;
        return head;
    }

    static void print(Node head) {
        System.out.print(head.index);
        while (head.next != null) {
            System.out.print(" " + head.next.index);
            head = head.next;
        }
        System.out.println();
    }

    static Node reverst(Node head) {
        if (head == null) {
            return null;
        }
        //初始化
        /**
         * head.next表示当前需要移到第一个位置的node,first.next表示新的链表
         */
        Node first = new Node();
        first.next = head;
        while (head.next != null) {
            Node next = head.next;
            head.next = next.next;
            next.next = first.next;
            first.next = next;
            //每次循环完,fist.next是新的表头
            // head.next表示需要移动第一个位置的node
        }
        //head.next为null,循环结束,表示已经移动完
        Node result = first.next;
        first.next = null;
        return result;
    }


    public static void main(String args[]) {
        Node head = init();
        print(head);
        head = reverst(head);
        print(head);
    }
}

证明

初始化

head.next表示当前需要移到第一个位置的node,first.next表示新的链表

循环

每次循环完，head.next表示还是当前移到第一个位置的node，first.next还是表示新的链表

结束

head.next为null,循环结束,表示已经移动完