[面试题06]从未到头打印链表(LeetCode-剑指Offer)

//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 
//


//leetcode submit region begin(Prohibit modification and deletion)

import java.util.ArrayList;
import java.util.List;

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        List<Integer> list = new ArrayList<>();
        while (head!=null){
            list.add(head.val);
            head = head.next;
        }

        int [] in = new int[list.size()];

        for (int i = list.size()-1; i >= 0 ; i--) {
            in[(list.size()-i-1)] = list.get(i);
        }

        return in;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

 static  public int[] reversePrint(ListNode head) {
        StringBuffer sb = new StringBuffer();

        while (head!=null){
            sb.append(head.val);
            head = head.next;
        }


        int [] in = new int[sb.length()];

//        for (int i =   sb.length()-1; i >= 0; i--) {
//            System.out.println("a="+sb.charAt(i)+"---->"+(sb.length()-i-1));
//        }

        for (int i = sb.length()-1; i >= 0 ; i--) {
            in[(sb.length()-i-1)] = Integer.valueOf(String.valueOf(sb.charAt(i)));
        }

        return in;
    }

public int[] reversePrint1(ListNode head) {
        String s = "";
        while (head!=null){

            s = s +head.val + ",";
            head = head.next;
        }

        s = s.substring(0,s.length()-1); //去掉最后一个空格。

        String[] split = s.split(",");
        for (int i = 0; i < split.length; i++) {

        }

        int [] in = new int[split.length];

        int j = 0;
        for (int i = split.length-1; i >= 0 ; i--) {
            in[j] = Integer.valueOf(split[i]);
            j++;
        }
        return in;
    }