题目描述

输入一个链表，按链表值从尾到头的顺序返回一个ArrayList。

法一：递归

include <iostream>

include <vector>

using namespace std;

struct ListNode
{
public:
int val;
struct ListNode *next;

/*ListNode(int x) :
val(x), next(NULL)
{
}*/

};

class Solution
{
private:
vector<int> res;

public:
vector<int> printListFromTailToHead(ListNode *head)
{
Recurse(head);
return res;
}

void Recurse(ListNode *head)
{
    if (head)
    {
        Recurse(head->next);
        res.push_back(head->val);
    }
}

};

int main()
{
ListNode list[4];
list[0].val = 1;
list[0].next = &list[1];
list[1].val = 2;
list[1].next = &list[2];
list[2].val = 3;
list[2].next = &list[3];
list[3].val = 4;
list[3].next = NULL;

Solution solu;
vector<int> res = solu.printListFromTailToHead(list);

cout << "there are " << res.size() << "datas in vector" << endl;
for (int i = 0; i < res.size(); i++)
{
    cout << res[i] << endl;
}
cout << system("pause");
return 0;

}

法二：利用栈“后进先出”的思路实现

include <iostream>

include <vector>

include <stack>

using namespace std;

struct ListNode
{
public:
int val;
struct ListNode *next;

/*ListNode(int x) :
val(x), next(NULL)
{
}*/

};

class Solution
{
private:
vector<int> res;

public:
vector<int> printListFromTailToHead(ListNode head)
{
stack<int> st;
ListNode cur = head;

    while (cur)
    {
        st.push(cur->val);
        cur = cur->next;
    }
    while(!st.empty())
    {
        res.push_back(st.top());
        st.pop();
    }
    return res;
}

};

int main()
{
ListNode list[4];
list[0].val = 1;
list[0].next = &list[1];
list[1].val = 2;
list[1].next = &list[2];
list[2].val = 3;
list[2].next = &list[3];
list[3].val = 4;
list[3].next = NULL;

Solution solu;
vector<int> res = solu.printListFromTailToHead(list);

cout << "there are " << res.size() << "datas in vector" << endl;
for (int i = 0; i < res.size(); i++)
{
    cout << res[i] << endl;
}
cout << system("pause");
return 0;

}

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参考资料：https://blog.csdn.net/gatieme/article/details/51107632