我完成了本周到期的项目。它是一个程序，按升序获取数组的交集和差异。一切都在方法的范围内发挥作用，但我试图在用户的反应之间加入一个线条，但我必须在某个地方删除某些东西并且不能使用。找到我的日志到那个部分。它假设只是重复do while循环中的内容，但是当我第二次再次进行时，看起来它似乎是将我设置的数字乘以int。我知道它必须是一个简单的逻辑错误，但我的头脑在这一点上被炸了。
这是我到目前为止所得到的：
import java.util.Scanner;
public class settest {
public static Scanner kbd = new Scanner(System.in);
/*
* This method will retrieve "size" number of unique
* integers from keyboard and placed into an array set.
*/
public static void getData(int [] set, int size){
boolean isUnique=false;
int input; int count=0;
while(count < size){
input = kbd.nextInt();
for(int i=0; i < size; i++)
{
if (input == set[i])
{
isUnique = true;
}
}
if (!isUnique)
{
set[count] = input;
count++;
}
isUnique = false;
}
}
/*
* This method will calculate the intersection of two sets of arrays.
* It needs to receive two sets of arrays, their size, and a holding
* array as parameters from the main program.
* When finished the method will return the numbers that are in common
* in both sets (resultSize)
*/
public static int intersection(int [] setA, int sizeA , int [] setB,
int sizeB, int [] resultSet){
int resultSize=0;
for(int x=0; x<sizeA; x++)
{
for(int y=0; y<sizeB; y++)
{
if(setA[x]==setB[y])
{
resultSet[resultSize]=setA[x];
resultSize++;
}
}
}
return resultSize;
}
/*
* Calculates the difference (A-B) of the two array sets.
* It will received the set [], size of correct array and in
* ascending order as parameters.
* Returns the correct array size for the received resultSet[]
*/
public static int difference(int [] setA, int sizeA , int [] setB,
int sizeB, int [] resultSet){
int resultSize=0; boolean sameVal=false;
for(int x=0; x<sizeA; x++ ){
for(int y=0; y<sizeB; y++){
if(setA[x]==setB[y]){
sameVal=true;
}
}
if(!sameVal){
resultSet[resultSize]=setA[x];
resultSize++;
}
sameVal=false;
}
return resultSize;
}
/*
* Method will sort the numbers in set[] array into ascending order.
*/
public static void sort(int [] set, int size){
for(int x=0; x< size-1; x++)
{
for(int y=x+1; y<size; y++)
{
if(set[x] > set[y])
{
int temp = set[x];
set[x] = set[y];
set[y] = temp;
}
}
}
}
public static final int MAXSIZE = 20;
public static void main(String[] args) {
int [] setA = new int[MAXSIZE];
int [] setB = new int[MAXSIZE];
int [] resultSet = new int[MAXSIZE];
int sizeA=0, sizeB=0;int resultSize=0;
String repeat;
do{
System.out.println("ENTER 2 SETS OF NUMBERS.\n");
System.out.println("HOW MANY NUMBERS IN YOUR FIRST SET?");
sizeA=kbd.nextInt();
for(int i=0; i<sizeA; i++){
if (sizeA>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED."
+ "ENTER A DIFFERENT AMOUNT. \n");
sizeA = kbd.nextInt();
}
}
System.out.println("\nENTER THE FIRST SET OF NUMBERS");
getData(setA, sizeA);
sort(setA,sizeA);
System.out.println("\nHOW MANY NUMBERS IN YOUR SECOND SET?");
sizeB=kbd.nextInt();
for(int i=0; i<sizeB; i++){
if (sizeB>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED. \n"
+ "ENTER A DIFFERENT AMOUNT.\n ");
sizeB = kbd.nextInt();
}
}
System.out.println("\nENTER THE SECOND SET OF NUMBERS");
getData(setB, sizeB);
sort(setB, sizeB);
System.out.println("\nINTERSECTION:");
resultSize=intersection(setA,sizeA,setB,sizeB, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i] + " ");
}
System.out.println("\n\nDIFFERENCE OF A-B:");
resultSize=difference(setA,sizeA,setB,sizeB, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i]+" ");
}
System.out.println("\n\nDIFFERENCE OF B-A:");
resultSize=difference(setB,sizeB,setA,sizeA, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i]+" ");
}
System.out.println("\n\nDO YOU WANT TO CONTINUE?");
repeat= kbd.next().toLowerCase();
} while(repeat.equals("y"));
}
}
任何帮助将不胜感激。
ENTER 2 SETS OF NUMBERS.
你的第一套中有多少个？ 3
输入第一组数字 4 3 2
你的第二套中有多少个？ 2
输入第二组数字 五 6
路口：
A-B的差异： 2 3 4
B-A的差异： 5 6
你想继续吗？ ÿ 输入2组数字。
你的第一套中有多少个？ 1
输入第一组数字 6
你的第二套中有多少个？ 6
输入第二组数字 3 五 4 1 0 9 7
路口：
A-B的差异： 6
B-A的差异： 1 3 4 5 7 9
你想继续吗？ 8 再见
在第二次循环中，用户数量以某种方式递增到广告1或一对
的位置答案 0 :(得分：1)
如果没有预览输出，很难说，但我会采取刺激措施。对于我来说，在do / while的第二次迭代期间，如果您使用的数组大小小于原始数组，那么您可以在第一次迭代中保留垃圾。
最好在每次迭代期间重新初始化这些变量。
do {
System.out.println("ENTER 2 SETS OF NUMBERS.\n");
System.out.println("HOW MANY NUMBERS IN YOUR FIRST SET?");
sizeA=kbd.nextInt();
while(sizeA>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED."
+ "ENTER A DIFFERENT AMOUNT. \n");
sizeA = kbd.nextInt();
}
System.out.println("\nENTER THE FIRST SET OF NUMBERS");
setA = new int[MAXSIZE];
getData(setA, sizeA);
sort(setA,sizeA);
System.out.println("\nHOW MANY NUMBERS IN YOUR SECOND SET?");
sizeB=kbd.nextInt();
while(sizeB>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED. \n"
+ "ENTER A DIFFERENT AMOUNT.\n ");
sizeB = kbd.nextInt();
}
System.out.println("\nENTER THE SECOND SET OF NUMBERS");
setB = new int[MAXSIZE];
getData(setB, sizeB);
sort(setB, sizeB);
System.out.println("\nINTERSECTION:");
.
.
.
.
} while (repeat.equal('y'))
这可以解决您的问题吗？就像我之前说过的那样，我并不完全理解你的问题，所以如果不解决这个问题，请提供更多细节，以便我能提供帮助。
答案 1 :(得分：1)
我确信以下代码会对您有所帮助。您可以将初始化数组部分移动到while循环中，以便可以使用用户输入初始化数组的大小。
public static void main(String[] args) {
int sizeA=0, sizeB=0;int resultSize=0;
String repeat;
do{
System.out.println("ENTER 2 SETS OF NUMBERS.\n");
System.out.println("HOW MANY NUMBERS IN YOUR FIRST SET?");
sizeA=kbd.nextInt();
int [] setA = new int[sizeA ]; //Initializing Array of size that user inputs.
for(int i=0; i<sizeA; i++){
if (sizeA>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED."
+ "ENTER A DIFFERENT AMOUNT. \n");
sizeA = kbd.nextInt();
}
}
System.out.println("\nENTER THE FIRST SET OF NUMBERS");
getData(setA, sizeA);
sort(setA,sizeA);
System.out.println("\nHOW MANY NUMBERS IN YOUR SECOND SET?");
sizeB=kbd.nextInt();
int [] setB = new int[sizeB]; //Initializing Array of size that user inputs.
for(int i=0; i<sizeB; i++){
if (sizeB>MAXSIZE){
System.out.println("\nERROR: SET SIZE LIMIT EXCEEDED. \n"
+ "ENTER A DIFFERENT AMOUNT.\n ");
sizeB = kbd.nextInt();
}
}
System.out.println("\nENTER THE SECOND SET OF NUMBERS");
getData(setB, sizeB);
sort(setB, sizeB);
/* Initializing the result array with Largest among sizeA and sizeB*/
resultSize = sizeA;
if(sizeB > sizeA)
{
resultSize = sizeB;
}
int [] resultSet = new int[resultSize];
System.out.println("\nINTERSECTION:");
resultSize=intersection(setA,sizeA,setB,sizeB, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i] + " ");
}
System.out.println("\n\nDIFFERENCE OF A-B:");
resultSize=difference(setA,sizeA,setB,sizeB, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i]+" ");
}
System.out.println("\n\nDIFFERENCE OF B-A:");
resultSize=difference(setB,sizeB,setA,sizeA, resultSet);
for(int i=0; i<resultSize; i++){
System.out.print(resultSet[i]+" ");
}
System.out.println("\n\nDO YOU WANT TO CONTINUE?");
repeat= kbd.next().toLowerCase();
} while(repeat.equals("y"));