问题
What is the best approach to solve the following problem in Java? There are 3 two-dimensional arrays that have equal number of rows and columns:
Array1:
[1]: {1, 2, 3}
[2]: {2, 2, 4}
[3]: {1, 1, 1}
Array2:
[1]: {1, 2, 0}
[2]: {1, 2, 3}
[3]: {1, 0, 1}
Array3:
[1]: {1, 1, 1}
[2]: {1, 2, 3}
[3]: {1, 0, 1}
I need to find out the indexes of rows that exist in all arrays. In the above example the answer should be: [1],[2],[2]
Array1 [1]: {1, 2, 3}
Array2: [2]: {1, 2, 3}
Array3: [2]: {1, 2, 3}
UPDATE: Is there any built-in function to do this? Or is the FOR loop the unique solution?
回答1:
I would do the following:
- Create a class that takes a row in the array and implements
hashcodeandequals. - Insert each row (wrapped in the above class) in the first two arrays into two
HashMaps - for each row in the last array, determine if they exist in the
HashMaps
Edit #2, realized the mapping would need to be reversed.
private Map<MyClass, Integer> map(int[] array){
Map<MyClass, Integer> arrayMap = new HashMap<>();
for (int i; i<array.length; i++)
arrayMap.put(new MyClass(array[i]), i);
}
private mySearch(){
int[] array1, array2, array3;
Map<MyClass, Integer> map1 = map(array1);
Map<MyClass, Integer> map2 = map(array2);
for (int i=0; i<array3.lenght; i++){
MyClass row = new MyClass(array3[i]);
Integer array1Row = map1.get(row);
Integer array2Row = map2.get(row);
if (array1Row != null && array2Row != null){
// Matching rows found
}
}
}
回答2:
AFAIK there is no built-in function for this.
You need to write your own function to implement it (using for loop).
post some code to show what have you tried, if it's not working corectly OR for optimization.
回答3:
Check this code:
ArrayList<Integer[]> arr1 = new ArrayList<Integer[]>();
arr1.add(new Integer[]{1,2,3});
arr1.add(new Integer[]{2,2,5});
arr1.add(new Integer[]{1,1,1});
ArrayList<Integer[]> arr2 = new ArrayList<Integer[]>();
arr2.add(new Integer[]{1,2,0});
arr2.add(new Integer[]{1,2,3});
arr2.add(new Integer[]{1,0,1});
ArrayList<Integer[]> arr3 = new ArrayList<Integer[]>();
arr3.add(new Integer[]{1,1,1});
arr3.add(new Integer[]{1,2,3});
arr3.add(new Integer[]{1,0,1});
for (int r = 0 ; r < arr1.length() ; r++) {
for(int s = 0 ; s < arr2.length() ; s++){
for(int t = 0 ; t < arr3.length() ; t++){
if(Arrays.deepEquals(arr1.get(r), arr2.get(s))){
}else
{
continue;
}
if(Arrays.deepEquals(arr1.get(r), arr3.get(t))){
System.out.println(r + " " + s + " " +t);;
}
}
}
}
回答4:
First of all, write a function called AreRowsEqual(), which compares two rows. So, now, the problem has been restated as Find similar elements in 3 arrays.
Secondly, try to think of a solution that would be the closest to something you already know based on prior knowledge: to find equal elements in two arrays, you need a double nested for loop. So, to find equal elements in three arrays, you would need a triple nested loop, right?
Okay, now, cross this out as a bad, bad, bad solution, because its time complexity is O(n^3). We should be able to do better.
Consider this: in order for an element to be similar in all 3 arrays, first it has to be similar among the first two; then, it has to be similar among the next two. The complexity of such an algorithm will be something akin to O(x*n), where x is the number of arrays. Much better, right? (I cannot figure out precisely what the O() will be, help anyone?) EDIT: it turns out it is O((n^2)*(x-1)), which is a lot better than O(n^3) when n > x --END EDIT This, incidentally, allows you to forget about the requirement for strictly 3 arrays and just consider a number of x arrays.
I wrote an approach which received one upvote and then I realized that it was not going to work and I deleted it. This is another try, which I believe will work:
Create a two dimensional array of integers. We will call this 'the matrix'. This matrix will have x columns, and the number of rows will be the number of rows of your first array. (Yes, this will work even if your arrays have differing lengths.) The numbers in the cells of this matrix will be matching row indexes. So, for example, after the algorithm I am describing finishes, a row of { 1, 3, 2 } in the matrix will tell us that row 1 of the first array matches row 3 of the second array and row 2 of the third array. We will use -1 to indicate 'no match'.
So, the first column of the matrix needs to be initialized with the indexes of all rows of your first array, that is, with the numbers 0, 1, 2, ... n where n is the number of elements in the first array.
For each additional array, fill its column in the matrix as follows: loop through each row of the current column in the matrix, and compute the cell as follows: if the corresponding cell of the previous column was -1, carry the -1 over into this cell. Otherwise, look for a row in the current array which matches the corresponding row of the previous array, and if found, store its index into this cell. Otherwise, store -1 in this cell.
Repeat for all of your arrays, that is, for all the columns in the matrix. In the end, your matching rows are those that do not have a -1 in the last column of the matrix.
If you really care about efficiency, you can do as John B suggested, and write an immutable class called Row which encapsulates (contains a reference to) a row and implements hashCode() and equals(). equals() here can be implemented using Arrays.deepEquals(). There may also be some goodie in Arrays called deepHashCode(), or else you will need to roll your own.
回答5:
Write a method that accepts two arguments: a two dimensional array to search for, a one dimensional array row. The method returns -1 if no match found, otherwise return the index of the matching row.
For each row of the first array: 2a. Call the method passing the row from first array and Array2. 2b. If 2a returns >0, call the method passing the same row and Array3
If 2b returns >0, output the returned indexes.
回答6:
Check Arrays.deepEquals() method. It will check two arrays are equal or not.
来源：https://stackoverflow.com/questions/8787989/find-similar-rows-in-3-two-dimensional-arrays