我应该编写一个程序,从txt
数组中的IntegerType
文件中读取整数。我已经创建了一个实现我的AnyType接口的IntegerType
类,您将在代码中看到。然后,我应该对数组进行排序,并研究在排序过程中进行了多少次比较和交换以研究效率,我知道这是O(N^2)
。我在代码中设置了断点,向我显示整数被读入我的String []数字数组。当我尝试将它们添加到我的IntegerType
数组时,它会命中while (scan.hasNext())
代码行并完全跳过我的for循环以将整数添加到我的数组中。任何人都有任何建议如何解决这个问题?谢谢你的时间。这是我的代码:
我的Sorting
课程:
public class Sorting {
public static void main(String[] args) throws IOException {
int type, sort;
Scanner read = new Scanner(System.in);
//Ask user for data type of input
System.out.println("Make selection by typing corresponding integer value and pressing Enter.");
System.out.println("Select type of input:");
System.out.println("1 = Integer 2 = String");
type = read.nextInt();
//Ask user for sorting algorithm desired
System.out.println("Select sorting algorithm to be used:");
System.out.println("1 = Insertion 2 = Selection 3 = Bubble");
sort = read.nextInt();
//Read in integer values from generated .txt files into corresponding integer arrays
Scanner scan = new Scanner(new File("descending.txt"));
String line = scan.nextLine();
String[] numbers = line.split(" ");
IntegerType[] worstCase = new IntegerType[numbers.length];
while (scan.hasNext()) {
for (int i = 0; i < worstCase.length; i++) {
worstCase[i] = new IntegerType(scan.nextInt());
}
}
scan = new Scanner(new File("random.txt"));
line = scan.nextLine();
numbers = line.split(" ");
IntegerType[] avgCase = new IntegerType[numbers.length];
while (scan.hasNext()) {
for (int i = 0; i < numbers.length; i++) {
avgCase[i] = new IntegerType(scan.nextInt());
}
}
scan = new Scanner(new File("ascending.txt"));
line = scan.nextLine();
numbers = line.split(" ");
IntegerType[] bestCase = new IntegerType[numbers.length];
while (scan.hasNext()) {
for (int i = 0; i < numbers.length; i++) {
bestCase[i] = new IntegerType(scan.nextInt());
}
}
if ((type == 1 || type == 2) && (sort == 1)) //Insertion Ascending
{
System.out.println("Insertion Sort / Ascending / Worst Case");
Sort.insertionSort(worstCase, worstCase.length);
System.out.println("Insertion Sort / Ascending / Average Case");
Sort.insertionSort(avgCase, avgCase.length);
System.out.println("Insertion Sort / Ascending / Best Case");
Sort.insertionSort(bestCase, bestCase.length);
}
}
}
我的Sort
课程:
public class Sort {
public static void insertionSort(AnyType[] list, int size) {
int compare = 0, swap = 0;
AnyType key;
for (int i = 1; i < size; i++) {
key = list[i];
int j = i - 1;
compare++;
if ((j > -1) && (list[j].isBetterThan(key))) {
list[j + 1] = list[j];
j--;
swap++;
}
list[j + 1] = key;
}
System.out.println("There were " + compare + " comparisons made.");
System.out.println("There were " + swap + " swaps made.");
}
}
我的AnyType
界面
public interface AnyType {
public boolean isBetterThan(AnyType datum);
}
我的IntegerType
班级
public class IntegerType implements AnyType {
private int number;
IntegerType() {
number = 0;
}
IntegerType(int i) {
number = i;
}
IntegerType(String s) {
number = Integer.parseInt(s);
}
public boolean isBetterThan(AnyType datum) {
return (this.number > ((IntegerType) datum).number);
}
public int toInteger() {
return number;
}
}
答案 0 :(得分:0)
我想该文件由一行中的空格分隔整数值组成。 无论如何,你应该使用Scanner.hasNextInt()方法进行检查。 所以问题出在这里(见评论):
//Read in integer values from generated .txt files into corresponding integer arrays
Scanner scan = new Scanner(new File("descending.txt"));
String line = scan.nextLine(); //reads already the whole line
String[] numbers = line.split(" "); //now a string array with all the values
//there is nothing left, everything is already stored in "line"/"numbers" variables
while (scan.hasNext()) {
//something
}
由于您现在事先没有多少整数值,因此您需要一个IntegerType值的动态数据结构(例如,一个ArrayList)。或者,您遍历数字数组并将每个字符串值转换为int:
Scanner scan = new Scanner(new File("descending.txt"));
String line = scan.nextLine();
String[] numbers = line.split("\\s+");
IntegerType[] worstCase = new IntegerType[numbers.length];
for(int i = 0; i < numbers.length; ++i)
worstCase[i] = new IntegerType(numbers[i]); //will work because IntegerType has a constructor which accepts a string
}
您的Sort方法的另一个注释:它们不需要数组的长度作为参数,您可以使用array.length(但这只是我猜的味道)。