两个数组的交集得到索引出界错误

Question

程序是：查找两个数组的交集(公共元素(。

输入：arr1 -> {10.45, 14.0, 18.35, 88.88, 54.10, 18.35}

arr2 -> {17.20, 13.30, 10.45, 18.35, 84.33, 13.30}

输出：10.45, 18.35

方法签名 ： double[] getIntersectionOfArray(double[] arr1， double[] arr2(

方法：删除两个数组的重复元素，比较两个数组并赋予第三个数组的公共值问题：一旦 m 调用第二个数组的方法，它就会给出索引出边界数组。 有人可以解释为什么会发生这种情况。

限制：不使用收集方法

import java.util.Arrays;

public class Test1 {
int count = 0;
int resultcount = 0;
void findIntersectionOfArray(double[] array1, double[] array2) {
double[] tempArray = new double[array1.length ]; //12

double[] tempArray1 = new double[array2.length];
// Calling Method to find Unique Array
tempArray = uniqueArray1(array1);
tempArray1 = uniqueArray1(array2);

// Removing 0.0 Extra value & Assigning array1 Unique value in array result 1
double[] resultArray1 = new double[count];
for (int index = 0; index < resultArray1.length; index++) {
resultArray1[index] = tempArray[index];
}
// Removing 0.0 Extra value & Assigning array 2 Unique value in array result 2
double[] resultArray2 = new double[count];
for (int index = 0; index < resultArray2.length; index++) {
resultArray2[index] = tempArray[index];
}
//System.out.println("Unique Array 1 : " + Arrays.toString(resultArray1));
//System.out.println("Unique Array 2 : " + Arrays.toString(resultArray2));
// Calling Method to Get Size of an intersection array
int size = getSizeOfAnArray(array1, array2);
double[] intersectArray = new double[size];
// Finding Common Elements between between 2 Arrays
boolean flag = true;
for (int outerindex = 0; outerindex < resultArray1.length; outerindex++) {
flag = true;
for (int innerindex = 0; innerindex < resultArray2.length; innerindex++) {
if (resultArray1[outerindex] == resultArray2[innerindex]) {
flag = false;
}
}
if (flag == false) {
intersectArray[outerindex] = resultArray1[outerindex];
count++;
}
} // O/p : [10.45, 0.0, 18.35, 0.0, 0.0, 0.0]
//System.out.println("Intersection array " + Arrays.toString(intersectArray));
// Printing Final Array Value getting 0.0 as well
for (int index = 0; index < intersectArray.length; index++) {
System.out.print(intersectArray[index] + " ,");
}
}
// Get Size of an intersectArray
int getSizeOfAnArray(double[] array1, double[] array2) {
int size = 0;
if (array1.length < array2.length)
size = array1.length;
else
size = array2.length;
return size;
}
// Method Unique Array will give Array1 Unique element
double[] uniqueArray1(double[] givenArray) {
double[] tempArray = new double[givenArray.length]; // 6
boolean isNumberPresent = true;
for (int outerindex = 0; outerindex < givenArray.length; outerindex++) {

isNumberPresent = true;
for (int innerindex = 0; innerindex < givenArray.length; innerindex++) {
if ((givenArray[outerindex] == tempArray[innerindex])) {
isNumberPresent = false;
}
}
if (isNumberPresent) {
tempArray[count] = givenArray[outerindex];
count++;
}
}
return tempArray; // [10.45, 14.0, 18.35, 88.88, 54.1,0.0]
}

public static void main(String[] args) {
Test1 test = new Test1();
double[] array1 = { 10.45, 14.0, 18.35, 88.88, 54.10, 18.35 };
double[] array2 = { 17.20, 13.30, 10.45, 18.35, 84.33, 13.30 };
test.findIntersectionOfArray(array1, array2);
}
}

Answer 1

• 我注意到的一个问题是您没有在uniqueArray1中重置count。我会在方法的开头将其设置为 0。
• 应将count值复制到retArray，如下所示。

double[] retArray = new double[count];
for (int i = 0; i < count; i++) {
retArray[i] = tempArray[i];
}
return retArray; 

这将摆脱任何挥之不去的 0.0 值。 正确的技术是以相同的方法完成此操作，以消除进一步处理的负担。

• 在下面的循环中，请参阅我的评论。

// need to reset count to 0
count = 0;
// Finding Common Elements between between 2 Arrays
boolean flag = true;
for (int outerindex = 0; outerindex < resultArray1.length;
outerindex++) {
flag = true;
for (int innerindex = 0; innerindex < resultArray2.length;
innerindex++) {
if (resultArray1[outerindex] == resultArray2[innerindex]) {
flag = false;  
// might as well get out of this loop since flag will never 
// be true.                    
break;
}
}
if (flag == false) {
// important to assign to location count (at the beginning of the
// array).
intersectArray[count] = resultArray1[outerindex];
count++;
}
}

• 打印值时，请使用计数，而不是数组长度。

for (int index = 0; index < count; index++) {
System.out.print(intersectArray[index] + " ,");
}

这是完整的程序。 有些名字略有变化。

import java.util.Arrays;
public class Test1 {

void findIntersectionOfArray(double[] array1, double[] array2) {
// get unique elements and return arrays.
double[] resultArray1 = uniqueArray1(array1); //12
double[] resultArray2 = uniqueArray1(array2);

// initialize temporary result array
double[] tempResultArray = new double[array1.length];
// Finding Common Elements between between 2 Arrays

boolean flag = true;
int count = 0;
for (int outerindex = 0; outerindex < resultArray1.length; outerindex++) {
flag = true;
for (int innerindex = 0; innerindex < resultArray2.length; innerindex++) {
if (resultArray1[outerindex] == resultArray2[innerindex]) {
flag = false;
break; // break out if false since it won't go back to true
}
}
// add element to temp result array at count location
if (flag == false) {
tempResultArray[count] = resultArray1[outerindex];
count++;
}
} 

// prepare to fill final result array of size count
double[] finalResultArray = new double[count];

for (int index = 0; index < count; index++) {
finalResultArray[index] = tempResultArray[index];
}
// print the array.
System.out.println(Arrays.toString(finalResultArray));
}
// Method Unique Array will give Array1 Unique element
double[] uniqueArray1(double[] givenArray) {
int count = 0;
double[] tempArray = new double[givenArray.length]; // 6
boolean isNumberPresent = true;
for (int outerindex = 0; outerindex < givenArray.length; outerindex++) {

isNumberPresent = true;
for (int innerindex = 0; innerindex < givenArray.length; innerindex++) {
if ((givenArray[outerindex] == tempArray[innerindex])) {
isNumberPresent = false;
}
}
if (isNumberPresent) {
tempArray[count] = givenArray[outerindex];
count++;
}
}

// only copy valid array elements
double[] returnArray = new double[count];
for (int i = 0; i < count; i++) {
returnArray[i] = tempArray[i];
}
return returnArray; 
}

public static void main(String[] args) {
Test1 test = new Test1();
double[] array1 = { 10.45, 14.0, 18.35, 88.88, 54.10, 18.35 };
double[] array2 = { 17.20, 13.30, 10.45, 18.35, 84.33, 13.30 };
test.findIntersectionOfArray(array1, array2);
}
}

Answer 2

我是否可以建议一种更简单的方法来解决该问题，例如 Set

1. 处理 array1 并将所有元素添加到名为 array1UniqSet 的集合中
2. 之后遍历 array2 中的每个元素，并检查 array1UniqSet 是否包含该元素： ->如果元素存在，则将它们添加到某个输出列表中 -> 如果不向前移动数组 1 中的下一个元素