“本周晚些时候我将进行的测试将不允许使用数组或任何花哨的实用程序，我们应该只使用扫描仪和循环。”
没有数据结构的排序会适得其反，但我离题了。
在这种情况下，我认为对数据进行排序的唯一方法是：一次读取一行数据，修剪该行，并将其附加到带有尾随空格的String。然后你将不得不想出一些算法来爬取String，并比较和交换它的sub-strings，直到它被排序。根据经验，这不是最令人愉快的活动。
这是我刚刚输入的一个工作示例。它使用一个Scanner 进行文件输入，并使用Strings。 Strings 不是数组，这很好。我试图通过代码中的 cmets 进行解释，但如果您需要更好的解释，请在此答案的 cmets 中告诉我。
注意：这个例子只是对输入进行排序，它没有找到最高的 2 个等级，但是一旦输入被排序，这应该很容易。只需将排序后的String 传递给Scanner，解析一些数据，然后瞧。
Driver.java：
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Driver {
public static void main(String[] args) {
Scanner in = null;
String inputStr = "";
try {
// open the file "xin.txt" to retrieve input
in = new Scanner(new File("xin.txt"));
} catch(FileNotFoundException e) {
System.out.println(e.getMessage());
System.exit(0);
}
// read the data into a one line string
while(in.hasNext()) {
inputStr += in.nextLine().trim() + " ";
}
System.out.println("[ORIGINAL]\n" + inputStr + "\n");
System.out.println("[Start Sorting]");
inputStr = sortString(inputStr);
System.out.println("[Done Sorting]\n");
System.out.println("[Sorted String]\n" + inputStr);
}
private static String sortString(String orig) {
String one = "";
String two = "";
int oneStart = 0; // starting index of string to compare with
int twoStart = 0; // starting index of string to compare against
int oneStop = 0; // ending index of string to compare with
int twoStop = 0; // ending index of string to compare against
int oneTot = 0; // length of string one plus the grades
int twoTot = 0; // length of string two plus the grades
int cp = 0; // current position (index)
boolean onOne = false; // current position is on one
boolean onTwo = false; // current position is on two
boolean passOne = false; // passed string one
boolean passTwo = false; // passed string two
boolean notDone = true; // control boolean
while(notDone) {
if(cp >= orig.length() || isAtoZ(orig.charAt(cp))) {
if(!onOne && !passOne) {
oneStart = cp;
onOne = true;
} else if(passOne && (!onTwo && !passTwo)) {
twoStart = cp;
onTwo = true;
// total length of the first string is the current position
// minus the start of the first string
oneTot = cp - oneStart;
} else if(passTwo) {
// total length of the second string is the current position
// minus the start of the second string
twoTot = cp - twoStart;
one = orig.substring(oneStart, oneStop);
two = orig.substring(twoStart, twoStop);
// output the results of the comparing
System.out.println(orig);
System.out.println("Comparing: " + one + " to " + two);
System.out.println("Result: " + one.compareTo(two) + "\n");
// if the first string is alphabetically larger, then swap
// and restart sort,else continue to the next comparison
if(one.compareTo(two) > 0) {
orig = rangeSwap(oneStart, oneStart + oneTot,
twoStart, twoStart + twoTot, orig);
onOne = false;
onTwo = false;
passOne = false;
passTwo = false;
cp = -1;
} else {
onOne = false;
onTwo = false;
passOne = false;
passTwo = false;
cp = twoStart - 1;
}
}
} else {
if(onOne && !passOne) {
oneStop = cp;
passOne = true;
onOne = false;
} else if(onTwo && !passTwo) {
twoStop = cp;
passTwo = true;
onTwo = false;
}
}
// increment the current position by one
cp = cp + 1;
// the last string has no string to compare to so
// set the control boolean
if(cp >= (orig.length()-1) && (!passTwo)) {
notDone = false;
}
}
// return the sorted string
return orig;
}
private static String rangeSwap(int ob, int oe, int tb, int te, String s) {
String start = "";
String x = "";
String middle = "";
String y = "";
String end = "";
start = s.substring(0, ob);
x = s.substring(ob, oe);
middle = s.substring(oe, tb);
y = s.substring(tb, te);
end = s.substring(te);
return start + y + middle + x + end;
}
private static boolean isAtoZ(char what) {
// ASCII character codes: A to Z = 65 to 90
// a to z = 97 to 122
return (((int)what >= 65 && (int)what <= 90) ||
((int)what >= 97 && (int)what <= 122 ));
}
}
xin.txt：
Nick 85 90 76 Hannah 86 91 66 Joe
22 35 100
Zak 100
40 80
输出：
[Original]
Nick 85 90 76 Hannah 86 91 66 Joe 22 35 100 Zak 100 40 80
[Start Sorting]
Nick 85 90 76 Hannah 86 91 66 Joe 22 35 100 Zak 100 40 80
Comparing: Nick to Hannah
Result: 6
Hannah 86 91 66 Nick 85 90 76 Joe 22 35 100 Zak 100 40 80
Comparing: Hannah to Nick
Result: -6
Hannah 86 91 66 Nick 85 90 76 Joe 22 35 100 Zak 100 40 80
Comparing: Nick to Joe
Result: 4
Hannah 86 91 66 Joe 22 35 100 Nick 85 90 76 Zak 100 40 80
Comparing: Hannah to Joe
Result: -2
Hannah 86 91 66 Joe 22 35 100 Nick 85 90 76 Zak 100 40 80
Comparing: Joe to Nick
Result: -4
Hannah 86 91 66 Joe 22 35 100 Nick 85 90 76 Zak 100 40 80
Comparing: Nick to Zak
Result: -12
[Done Sorting]
[Sorted String]
Hannah 86 91 66 Joe 22 35 100 Nick 85 90 76 Zak 100 40 80