方法一：辅助栈：时间O(n)，空间O(n)
题解：利用栈的先进后出特性，先遍历链表进行入栈，随后出栈并把元素放入数组
class Solution {
public:
vector<int> reversePrint(ListNode* head)
{
if (head == NULL)
return vector<int>();
vector<int> res;
stack<int> stc;
while (head != NULL)
{
stc.push(head->val);
head = head->next;
}
while (!stc.empty())
{
res.push_back(stc.top());
stc.pop();
}
return res;
}
};
方法二：递归：时间O(n)，空间O(n)
题解：在回溯的时候，添加元素到数组中
class Solution {
public:
vector<int> res;
void dfs(ListNode* root)
{
if (root == NULL)
return;
dfs(root->next);
res.push_back(root->val);
}
vector<int> reversePrint(ListNode* head)
{
dfs(head);
return res;
}
};
方法三：逆置链表：时间O(n)，空间O(n)
题解：先逆置链表，再遍历链表插入数据
class Solution {
public:
vector<int> reversePrint(ListNode* head)
{
if (head == NULL)
return vector<int>();
ListNode* cur = head;
ListNode* pre = head->next;
cur->next = NULL;
while (pre != NULL)
{
cur = pre;
pre = pre->next;
cur->next = head;
head = cur;
}
vector<int> res;
while (head != NULL)
{
res.push_back(head->val);
head = head->next;
}
return res;
}
};