思路
思路比较简单,先把整个链表跑一遍,统计元素个数,再把链表跑一遍,把数倒着装到数组里,返回数组即可
代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
ListNode* tmp=head;
vector<int>ans;
int num=0;
while(tmp)
{
num++;
tmp=tmp->next;
ans.push_back(0);
}
tmp=head;
while(tmp)
{
ans[--num]=tmp->val;
tmp=tmp->next;
}
return ans;
}
};