输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。
示例 1:
输入:head = [1,3,2]
输出:[2,3,1]
限制:
0 <= 链表长度 <= 10000
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
stack<int> s;
while(head)
{
s.push(head->val);
head = head->next;
}
vector<int> ans;
while(!s.empty())
{
ans.push_back(s.top());
s.pop();
}
return ans;
}
};
class Solution {
vector<int> ans;
public:
vector<int> reversePrint(ListNode* head) {
dfs(head);
return ans;
}
void dfs(ListNode* head)
{
if(!head)
return;
dfs(head->next);
ans.push_back(head->val);
}
};
class Solution {
vector<int> ans;
public:
vector<int> reversePrint(ListNode* head) {
if(!head)
return {};
head = reverseList(head);
while(head)
{
ans.push_back(head->val);
head = head->next;
}
return ans;
}
ListNode* reverseList(ListNode* head)
{ //反转链表,返回新表头
ListNode *prev = NULL, *nt = head->next;
while(head && head->next)
{
head->next = prev;
prev = head;
head = nt;
nt = nt->next;
}
head->next = prev;
return head;
}
};