从尾到头打印链表

题目描述

输入一个链表，从尾到头打印链表每个节点的值。

输入描述:

输入为链表的表头

输出描述:

输出为需要打印的“新链表”的表头

解法一：递归实现（效率低）

import java.util.ArrayList;

class ListNode{

    int val;
    ListNode next = null;

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

}

public class Solution {    

    static ArrayList<Integer> arrayList = new ArrayList<Integer>();

    public static ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        if (listNode != null) {
            printListFromTailToHead(listNode.next);
            arrayList.add(listNode.val);
        }
        return arrayList;
    }

    public static void main(String[] args) {

        ListNode a = new ListNode(1);
        ListNode b = new ListNode(2);
        ListNode c = new ListNode(3);
        a.next = b;
        b.next = c;


        System.out.println(printListFromTailToHead(a));

    }

}

解法

二：借助栈实现复杂度为O(n)

class ListNode{

    int val;
    ListNode next = null;

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

}


public class Solution {

    public static ArrayList<Integer> printListFromTailToHead(ListNode1 listNode) {

        Stack<Integer> stack = new Stack<Integer>();
        ArrayList<Integer> arry = new ArrayList<Integer>();

        while (listNode != null) {
            stack.push(listNode.val);
            listNode = listNode.next;
        }

        while (!stack.empty()) {
            arry.add(stack.pop());
        }

        return arry;
    }

    public static void main(String[] args) {
        ListNode1 a = new ListNode1(1);
        ListNode1 b = new ListNode1(2);
        ListNode1 c = new ListNode1(3);

        a.next = b;
        b.next = c;

        System.out.println(printListFromTailToHead(a));

    }

}