剑指 Offer 06. 从尾到头打印链表🌟🌟🌟🌟🌟简单
课后作业
问题描述
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
代码实现
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
if(head == null){
return new int[0];
}
int n = 0;
ListNode temp = head;
while(temp != null){
n++;
temp = temp.next;
}
int res[] = new int[n];
while(head != null){
res[--n] = head.val;
head = head.next;
}
return res;
}
}
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reversePrint(self, head):
"""
:type head: ListNode
:rtype: List[int]
"""
if head == None:
return []
n = 0
temp = head
while temp != None:
n += 1
temp = temp.next
res = [0]*n
while head != None:
n -= 1
res[n] = head.val
head = head.next
return res
C++
/**
* 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) {
if (head == NULL) {
return vector<int>();
}
int n = 0;
ListNode* temp = head;
while (temp != NULL) {
n++;
temp = temp->next;
}
vector<int> res(n);
while (head != NULL) {
n--;
res[n] = head->val;
head = head->next;
}
return res;
}
};
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reversePrint(head *ListNode) []int {
if head == nil {
return []int{}
}
n := 0
temp := head
for temp != nil {
n++
temp = temp.Next
}
res := make([]int, n)
for head != nil {
n--
res[n] = head.Val
head = head.Next
}
return res
}