剑指offer-26-从尾到头打印链表

原题：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/

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

思路：

方法一：递归

	int[] r;
    int idx = 0;
    public int[] reversePrint(ListNode head) {
        if (head == null) {
            return new int[]{};
        }
        reverse(head, 0); 
        return r;
    }
    private void reverse(ListNode head, int c) {
        if (head.next == null) {
            r = new int[c + 1];
            r[idx++] = head.val;
            return;
        }
        reverse(head.next, ++c);
        r[idx++] = head.val;
    }

方法二：先遍历，获得长度，然后再次遍历，将链表数据从数组尾部往头部放

/**
 * 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[]{};
        }
        ListNode cur = head;
        int c = 0;
        while (cur.next != null) {
            cur = cur.next;
            c++;
        }
        int[]  r = new int[c + 1];
        cur = head;
        for (int i = c; i >= 0; i--) {
            r[i] = cur.val;
            cur = cur.next;
        }
        return r;
    }
}