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 gIt 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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