Finding adjacent elements in a 2d array and counting them.
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2
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Im stumped on what to do for this part of my homework and could really use some help. I need to cycle through a given 2d array and find all similar elements that are adjacent to another and count that so for example
AA--B
AA--B
-AA--
----C
So the count would be 3 one for the As one for the Bs and one for the C, I just kinda need an idea where to start So far i have
public static int howManyOrganisms(char image){
int count = 0;
for (int i = 0; i < image.length; i++) {
for (int j = 0; j < image[i].length; j++) {
if(image[i][j] != '-') {
count++;
}
System.out.println();
}
return howManyOrganisms(image, count);
}
}
I need help figuring out how to track the total number of elements that are within contact of one another (so left, right, down, up) being another similar element.
java arrays for-loop recursion multidimensional-array
asked Nov 19 at 20:15
Forrest Walker
304
add a comment |
up vote
2
down vote
favorite
Im stumped on what to do for this part of my homework and could really use some help. I need to cycle through a given 2d array and find all similar elements that are adjacent to another and count that so for example
AA--B
AA--B
-AA--
----C
So the count would be 3 one for the As one for the Bs and one for the C, I just kinda need an idea where to start So far i have
public static int howManyOrganisms(char image){
int count = 0;
for (int i = 0; i < image.length; i++) {
for (int j = 0; j < image[i].length; j++) {
if(image[i][j] != '-') {
count++;
}
System.out.println();
}
return howManyOrganisms(image, count);
}
}
I need help figuring out how to track the total number of elements that are within contact of one another (so left, right, down, up) being another similar element.
java arrays for-loop recursion multidimensional-array
asked Nov 19 at 20:15
Forrest Walker
304
2
You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).
– paulsm4
Nov 19 at 20:22
1
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Im stumped on what to do for this part of my homework and could really use some help. I need to cycle through a given 2d array and find all similar elements that are adjacent to another and count that so for example
AA--B
AA--B
-AA--
----C
So the count would be 3 one for the As one for the Bs and one for the C, I just kinda need an idea where to start So far i have
public static int howManyOrganisms(char image){
int count = 0;
for (int i = 0; i < image.length; i++) {
for (int j = 0; j < image[i].length; j++) {
if(image[i][j] != '-') {
count++;
}
System.out.println();
}
return howManyOrganisms(image, count);
}
}
I need help figuring out how to track the total number of elements that are within contact of one another (so left, right, down, up) being another similar element.
java arrays for-loop recursion multidimensional-array
asked Nov 19 at 20:15
Forrest Walker
304
Im stumped on what to do for this part of my homework and could really use some help. I need to cycle through a given 2d array and find all similar elements that are adjacent to another and count that so for example
AA--B
AA--B
-AA--
----C
So the count would be 3 one for the As one for the Bs and one for the C, I just kinda need an idea where to start So far i have
public static int howManyOrganisms(char image){
int count = 0;
for (int i = 0; i < image.length; i++) {
for (int j = 0; j < image[i].length; j++) {
if(image[i][j] != '-') {
count++;
}
System.out.println();
}
return howManyOrganisms(image, count);
}
}
I need help figuring out how to track the total number of elements that are within contact of one another (so left, right, down, up) being another similar element.
java arrays for-loop recursion multidimensional-array
java arrays for-loop recursion multidimensional-array
asked Nov 19 at 20:15
Forrest Walker
304
asked Nov 19 at 20:15
Forrest Walker
304
asked Nov 19 at 20:15
Forrest Walker
304
asked Nov 19 at 20:15
Forrest Walker
304
asked Nov 19 at 20:15
Forrest Walker
304
304
2
You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).
– paulsm4
Nov 19 at 20:22
1
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33
add a comment |
2
You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).
– paulsm4
Nov 19 at 20:22
1
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33
2
2
You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:
howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).– paulsm4
Nov 19 at 20:22
You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:
howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).– paulsm4
Nov 19 at 20:22
1
1
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33
add a comment |
1 Answer
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up vote
1
down vote
In each iteration, you can use your i,j variables to "navigate" the 2d plane and see if any interacting items are the same. In each iteration you would check the following indexes to see if they are the same:
- image[i-1][j] (one row up)
- image[i+1][j] (one row down)
- image[i][j-1] (one left)
- image[i][j+1] (one right)
Of course for all of these statements first you should check if +1/-1 is still within the size of your matrix, otherwise you will end up with out of bounds exception.
answered Nov 19 at 20:24
peterxz
104110
add a comment |
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up vote
1
down vote
In each iteration, you can use your i,j variables to "navigate" the 2d plane and see if any interacting items are the same. In each iteration you would check the following indexes to see if they are the same:
- image[i-1][j] (one row up)
- image[i+1][j] (one row down)
- image[i][j-1] (one left)
- image[i][j+1] (one right)
Of course for all of these statements first you should check if +1/-1 is still within the size of your matrix, otherwise you will end up with out of bounds exception.
answered Nov 19 at 20:24
peterxz
104110
add a comment |
up vote
1
down vote
In each iteration, you can use your i,j variables to "navigate" the 2d plane and see if any interacting items are the same. In each iteration you would check the following indexes to see if they are the same:
- image[i-1][j] (one row up)
- image[i+1][j] (one row down)
- image[i][j-1] (one left)
- image[i][j+1] (one right)
Of course for all of these statements first you should check if +1/-1 is still within the size of your matrix, otherwise you will end up with out of bounds exception.
answered Nov 19 at 20:24
peterxz
104110
add a comment |
up vote
1
down vote
up vote
1
down vote
In each iteration, you can use your i,j variables to "navigate" the 2d plane and see if any interacting items are the same. In each iteration you would check the following indexes to see if they are the same:
- image[i-1][j] (one row up)
- image[i+1][j] (one row down)
- image[i][j-1] (one left)
- image[i][j+1] (one right)
Of course for all of these statements first you should check if +1/-1 is still within the size of your matrix, otherwise you will end up with out of bounds exception.
answered Nov 19 at 20:24
peterxz
104110
In each iteration, you can use your i,j variables to "navigate" the 2d plane and see if any interacting items are the same. In each iteration you would check the following indexes to see if they are the same:
- image[i-1][j] (one row up)
- image[i+1][j] (one row down)
- image[i][j-1] (one left)
- image[i][j+1] (one right)
Of course for all of these statements first you should check if +1/-1 is still within the size of your matrix, otherwise you will end up with out of bounds exception.
answered Nov 19 at 20:24
peterxz
104110
answered Nov 19 at 20:24
peterxz
104110
answered Nov 19 at 20:24
peterxz
104110
answered Nov 19 at 20:24
peterxz
104110
104110
add a comment |
add a comment |
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You're on the right track: 1) GOAL: count #/adjacent elements. 2) Define a function. EXAMPLE:
howManyOrganisms(). 3) Create some loops to examine every column in every row. 4) For each element, check up, down, right and left. If adjacent to one or more, then add to count. 5) Optimize (do you need to check "up" for 1st row, or "right" for last column, etc).– paulsm4
Nov 19 at 20:22
1
Seems like a classis 'flood fill' algorithm modification. Check this
– Victor Gubin
Nov 19 at 20:33