如何复制数组中的元素但不在第三个数组中的另一个数组中

Question

i have two arrays filled with random int values.What i want to do is put the elements that are in one of the arrays but not in the other on a third array.How can i do this?

int[] array1 = new int[10];
int[] array2 = new int[10];

for(int j=0;j<array2.length;j++){
    int alea = rant.nextInt(10);
    valorpos = alea;
    array1[j] = rant.nextInt(10);
    array2[j] = rant.nextInt(10);
}

can anybody help me to continue the code jus to achieve whatever i said,thank you very much what i want to achieve is:

array1:[5,7,9,2,9,8,3,2,4,8];
Array 2: [6,2,5,8,3,0,5,9,2,0];
resultat:[7,4,6,0,0]

Answer 1

The most straightforward way would be to use collections to do this. Create a list from the first array and a set from the second. Then remove all elements of the second collection from the first one. Finally, produce an array. Something along the lines:

List<Integer> list1 = Arrays.stream(array1).boxed().collect(Collectors.toList());
Set<Integer> set2 = Arrays.stream(array2).boxed().collect(Collectors.toSet());
list1.removeAll(set2);
Integer[] array3 = new Integer[list1.size()];
list1.toArray(array3);

Answer 2

You could try a nested for loop? Not the most efficient way but simple enough for your needs...

boolean contain;

//loop through the first array
for(int number1 : array1){
    //set contain to false
    contain = false;

    //loop through the second array
    for(int number2 : array2){

        //if number1 in first array equals number2 in second array
        if(number1 == number2){
            //set contain to true and break the loop
            contain = true;
            break;
        }
    }
    //if contain is still false
    if(!contain){
        //add number1 to third array...
    }
}

Answer 3

The best thing to do here is step back and think about what you're trying to solve. You have two arrays each with random numbers in, lets say they are something like:

Array A :

1,3,4,5,6

Array B :

4,9,4,5,6

So we need to find the the values within array A which are not in array B . How would you work this out on paper? You would start with the first number of array A and scan through each number within array B to check whether its in there, if it is we stop looking, else we continue to the last number. So for the example above we would find:

Array Result :

1,3

So lets translate this into some code:

// a list which will be used to put the unique values in
List<Integer> result = new ArrayList<>();

// loop through each number in the first array
for(int i=0; i< arrayOne.length; i++){

    boolean isInOtherArray = false;
    // check this number against each number in the second array
    for(int j=0; j<arrayTwo.length; j++){

        // if the number matches, then we know its not unique 
        // so we stop the search
        if(arrayOne[i] == arrayTwo[j]){
            isInOtherArray=true;
            break;
        }
    }
    // if its not in the other array we know its unique so it goes 
    // in our result list
    if(!isInOtherArray){
        result.add(arrayOne[i]);
    }
}

Answer 4

Try with two for loops. First loop through the first array and consider each current element, then loop through the second array to check if the current number from array1 is in array2 . If it is not present in array2 , then add it to array3 . Lastly I added a piece of code to trim array3 to size.

    int[] array1=new int[10];
    int[] array2=new int[10];
    int[] array3=new int[10];
    // populate array1 and array2 ....

    boolean isInArray2;
    int current;
    int counter = 0;

    for (int i = 0; i < array1.length; i++) {
        current = array1[i];
        isInArray2= false;
        for (int j = 0; j < array2.length; j++) {
            if(array2[j] == current) {
                isInArray2= true;
                break;
            }
        }

        if(!isInArray2) {
            // not in array2 -> must add in array3
            array3[counter] = current;
            counter++;
        }
    }

    // Trim array3 to avoid null values:
    int[] tmp = array3;
    array3=new int[counter];
    for (int i = 0; i < array3.length; i++) {
        array3[i] =tmp[i];
    }