问题
I need help writing a method to find out a sum of 2 different sized 2d arrays.
public static int[][] summary(int[][] tab1, int[][] tab2, int x) {
int[][] finalTab = new int[4][5]; // I took sizes of bigger one
if (x < 0) {
for (int i = 0; i < finalTab.length - 1; i++) {
for (int j = 0; j < finalTab[i].length - 1; j++) {
finalTab[i][j] = tab1[i][j] + tab2[i][j];
if (tab1[i][j] == 0) {
finalTab[i][j] = tab2[i][j];
}
}
}
for (int i = 0; i < finalTab.length; i++) {
for (int j = 0; j < finalTab[i].length; j++) {
System.out.print(" " + finalTab[i][j] + " ");
}
System.out.println();
}
}
return finalTab;
}
Input is:
int [][] tab1 = {
{2, 4, 6, 8},
{2, 4, 6, 8},
{2, 4, 6, 8},
};
int [][] tab2 = {
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
};
Output is:
3 7 11 15 0
3 7 11 15 0
3 7 11 15 0
0 0 0 0 0
Output should be:
3 7 11 15 9
3 7 11 15 9
3 7 11 15 9
1 3 5 7 9
How can I replace all of 0
with numbers corresponding to the index of tab2?
回答1:
you can do it more generic
public static int[][] summary(int[][] tab1, int[][] tab2, int x) {
int maxLenX = tab1.length > tab2.length ? tab1.length : tab2.length;
int maxLenY = tab1[0].length > tab2[0].length ? tab1[0].length : tab2[0].length;
int[][] finalTab = new int[maxLenX][maxLenY]; // i took sizes of bigger one
if (x < 0) {
for (int i = 0; i <= finalTab.length - 1; i++) {
for (int j = 0; j <= finalTab[i].length - 1; j++) {
if (i > tab1.length - 1 || j > tab1[i].length - 1) {
finalTab[i][j] = tab2[i][j];
} else if (i > tab2.length - 1 || j > tab2[i].length - 1) {
finalTab[i][j] = tab1[i][j];
} else {
finalTab[i][j] = tab1[i][j] + tab2[i][j];
}
}
}
for (int i = 0; i < finalTab.length; i++) {
for (int j = 0; j < finalTab[i].length; j++) {
System.out.print(" " + finalTab[i][j] + " ");
}
System.out.println();
}
}
return finalTab;
}
so you can call it wether like that
summary(tab2, tab1, -1);
or
summary(tab1, tab2, -1);
回答2:
This is a more straight-forward solution:
for (int i = 0; i < finalTab.length; i++)
for (int j = 0; j < tab1[i].length; j++) {
int v1 = (i<tab1.length && j<tab1[i].length) ? tab1[i][j] : 0;
int v2 = (i<tab2.length && j<tab2[i].length) ? tab2[i][j] : 0;
finalTab[i][j] = v1 + v2;
}
回答3:
I made it this way:
for (int i = 0; i < finalTab.length; i++)
for (int j = 0; j < finalTab[i].length; j++)
if (tab1.length > tab2.length) {
finalTab[i][j] = tab1[i][j];
} else
finalTab[i][j] = tab2[i][j];
for (int i = 0; i < tab1.length; i++) {
for (int j = 0; j < tab1[i].length; j++) {
finalTab[i][j] = tab1[i][j] + tab2[i][j];
}
}
回答4:
You can use IntStream.of(int...) method to concatenate two elements from different sources:
int[][] tab1 = {
{2, 4, 6, 8},
{2, 4, 6, 8},
{2, 4, 6, 8}};
int[][] tab2 = {
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9}};
int[][] tab3 = IntStream
// iterate over indices of tab2,
// because it is biggest array
.range(0, tab2.length).mapToObj(i ->
IntStream.range(0, tab2[i].length).map(j ->
// for each cell concatenate element
// from tab2 with element from tab1
// if exists, or with 0 otherwise
IntStream.of(tab2[i][j],
i < tab1.length && j < tab1[i].length ?
tab1[i][j] : 0)
// sum of two
// elements
.sum())
.toArray())
.toArray(int[][]::new);
Arrays.stream(tab3).map(Arrays::toString).forEach(System.out::println);
// [3, 7, 11, 15, 9]
// [3, 7, 11, 15, 9]
// [3, 7, 11, 15, 9]
// [1, 3, 5, 7, 9]
See also:
• Adding up all the elements of each column in a 2d array
• Rotating an int Array in Java using only one semicolon
来源:https://stackoverflow.com/questions/65373009/sum-of-2-different-2d-arrays