"Sword Finger Offer" 05: Print the linked list from end to beginning

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首先学习计算机科学及理论。接着形成自己编程的风格。然后把这一切都忘掉，尽管改程序就是了。—— 小浩

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Print linked list from end to beginning

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Title description

Enter a linked list and return an ArrayList in the order of the linked list values ​​from end to beginning.

Original title

solution

Solution 1 [Recommended]
Traverse the linked list, push the value of each linked list node into the stack, and finally pop the elements in the stack into the list in turn.

/**
*    public class ListNode {
*        int val;
*        ListNode next = null;
*
*        ListNode(int val) {
*            this.val = val;
*        }
*    }
*
*/
import java.util.ArrayList;
import java.util.Stack;

public class Solution {
    /**
     * 从尾到头打印链表
     * @param listNode 链表头节点
     * @return list
     */
    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        ArrayList<Integer> res = new ArrayList<>();
        if (listNode == null) {
            return res;
        }
        Stack<Integer> stack = new Stack<>();
        while (listNode != null) {
            stack.push(listNode.val);
            listNode = listNode.next;
        }
        while (!stack.isEmpty()) {
            res.add(stack.pop());
        }

        return res;
    }
}

Solution two [not recommended]
Use recursion:

If it is not the end node of the linked list, continue the recursion;
if it is, add it to the list.
This method is not recommended. When there are too many recursive layers, Stack Overflow is prone to occur.

/**
*    public class ListNode {
*        int val;
*        ListNode next = null;
*
*        ListNode(int val) {
*            this.val = val;
*        }
*    }
*
*/
import java.util.ArrayList;
import java.util.Stack;

/**
 * @author bingo
 * @since 2018/10/28
 */
public class Solution {
    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        ArrayList<Integer> res = new ArrayList<>();
        if (listNode == null) {
            return res;
        }

        addElement(listNode, res);
        return res;

    }

    private void addElement(ListNode listNode, ArrayList<Integer> res) {
        if (listNode.next != null) {
            // 递归调用
            addElement(listNode.next, res);
        }
        res.add(listNode.val);
    }
}

Idea expansion

Test case

1. Functional test (the input linked list has multiple nodes; the input linked list has only one node);
2. Special input test (input linked list node pointer is empty).

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