每日几道leetcode刷刷题!
传送门
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]
Python版
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
stack = []
while head:
stack.append(head.val)
head = head.next
return stack[::-1]
C++版
/**
* 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) {
vector<int> res;
stack<int> s;
while (head){
s.push(head->val);
head = head->next;
}
while (!s.empty()){
res.push_back(s.top());
s.pop();
}
return res;
}
};
st。s.empty()作为判断容器是否为空的函数top()函数的作用是:返回栈顶元素的值。(只是返回,但不弹出)