从文件输入（查找链接列表的第n个元素） [英] Input from a file (Find the nth element for the Linked List)

问题描述

问题是在链表中找到第n个元素，我找到了找到元素的问题。

但是我必须从文件中读取一个输入并从列表中输出第n个元素

例如

 输入看起来像这样
abcd 4 
efgh 2 
 
输出会像
a 
g 
  

据说第一个参数是一个路径包含一系列空格分隔字符的文件名，后跟一个表示列表中索引的整数（1为基础），每行一个

我不知道该怎么走对这个？我会首先读取文件并将每行存储在列表中？

解决方案

是的，您应该逐行读取文件，并将每行存储在 List<串GT; 。
然后，只要这样做：

  String line; //使用列表中的迭代器来获取每行
 String [] elements = line.trim（）。split（\\s）; 
 char [] chars = new char [elements.length  -  1]; 
 int index = Integer.parseInt（elements [elements.length  -  1]）; 
 for（i = 0; i< elements.length  -  1; i ++）
 char [i] = elements [i] .charAt（0）; 
  

现在你有一个数组中的字符了......而且你说你已经把剩下的东西弄清楚了。

The problem is to find nth element in a linked list, i have the problem figured out on finding the element.

But i have to read an input from a file and output the nth element from the list

For example

Input would look like this 
a b c d 4
e f g h 2

Output would like
a
g

It is stated "The first argument will be a path to a filename containing a series of space delimited characters followed by an integer representing a index into the list (1 based), one per line"

I am not sure how i would go about this? Would i first the read the file and store each line in a List?

解决方案

Yes, you should read the file line by line, and store each line in a List<String>. Then, simply do this:

String line; // use an iterator on the list to get each line
String[] elements = line.trim().split("\\s");
char[] chars = new char[elements.length - 1];
int index = Integer.parseInt(elements[elements.length - 1]);
for (i = 0; i < elements.length - 1; i++)
    char[i] = elements[i].charAt(0);

Now you have the characters in an array ... and you said you already have the rest figured out.

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