题目

剑指 Offer 06. 从尾到头打印链表

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：

1 2	输入：head = [1,3,2] 输出：[2,3,1]

题解

python

还是很好写的。很多数据结构的变换都不需要考虑。

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def reversePrint(self, head):
        res = []
        root = head
        while root:
            res.insert(0,root.val)
            root = root.next
        return res

if __name__ == '__main__':
    L = [1,2,3,4,5]
    Head = ListNode(1)
    R = Head
    i = 1
    while i < len(L):
        R.next = ListNode(L[i])
        i += 1
        R = R.next
    a = Solution()
    print(a.reversePrint(Head))

Java

这里用到了Stack类

package Slash_offer.easy;

import java.util.Stack;

//Definition for singly-linked list.
class ListNode {
      int val;
      ListNode next;
      ListNode(int x) {
          val = x;
      }
}
public class Slash_offer_06_从尾到头打印链表 {
    public static void main(String[] args) {
        Slash_offer_06_从尾到头打印链表 slash_offer_06_从尾到头打印链表 = new Slash_offer_06_从尾到头打印链表();
        int [] L = new int[]{1,2,3,4,5};
        ListNode Head = new ListNode(1);
        ListNode R = Head;
        int i = 1;
        while (i<L.length){
            R.next = new ListNode(L[i++]);
            R = R.next;
        }
        for (int i1 : slash_offer_06_从尾到头打印链表.reversePrint(Head)) {
            System.out.println(i1);
        }
    }
    public int[] reversePrint(ListNode head) {
        Stack<Integer> integers = new Stack<>();
        ListNode R = head;
        while (R!=null){
            integers.push(R.val);
            R = R.next;
        }
        int length = integers.size();
        int[] ints = new int[length];
        for (int i = 0; i < length ; i++) {
            ints[i] = integers.pop();
        }
        return ints;
    }
}