Mixing
permutation algorithm in java
I was trying to permutate using java and i noticed my code works for strings but the same algorithm does not work correctly on Arrays
here's the code for strings
public static void Recurse(String s,int i)
{
for(int j=i;j<(s.length());j++)
{
s=Swap(s,i,j);
if(i<(s.length()-1))
{
Recurse(s,i+1);
}
if(i==(s.length()-1))
{
Display(s);
}
}
}
public static String Swap(String newString,int i,int j)
{
char newChar=newString.toCharArray();
char temp=newChar[i];
newChar[i]=newChar[j];
newChar[j]=temp;
return String.valueOf(newChar);
}
public static void Display(String s)
{
System.out.println(s);
}
public static void main(String args)
{
String i=new String("123");
Recurse(i,0);
}
output:
123
132
213
231
312
321
works fine.
then here is the code for arrays
public static void Recurse(int newArray,int i)
{
for(int j=i;j<newArray.length;j++)
{
Swap(newArray,i,j);
if(i<(newArray.length-1))
{
Recurse(newArray,i+1);
}
if(i==(newArray.length-1))
{
Display(newArray);
}
}
}
public static void Display(int Array)
{
for(int i: Array)
{
System.out.print(i);
}
System.out.println();
}
public static void main(String args)
{
Recurse(new int {1,2,3},0);
}
output
123
132
312
321
123
132
Does not work correctly,So i was wondering what went wrong here.
java arrays string permutation
asked Nov 20 '18 at 17:18
user9169047user9169047
183
|
show 1 more comment
I was trying to permutate using java and i noticed my code works for strings but the same algorithm does not work correctly on Arrays
here's the code for strings
public static void Recurse(String s,int i)
{
for(int j=i;j<(s.length());j++)
{
s=Swap(s,i,j);
if(i<(s.length()-1))
{
Recurse(s,i+1);
}
if(i==(s.length()-1))
{
Display(s);
}
}
}
public static String Swap(String newString,int i,int j)
{
char newChar=newString.toCharArray();
char temp=newChar[i];
newChar[i]=newChar[j];
newChar[j]=temp;
return String.valueOf(newChar);
}
public static void Display(String s)
{
System.out.println(s);
}
public static void main(String args)
{
String i=new String("123");
Recurse(i,0);
}
output:
123
132
213
231
312
321
works fine.
then here is the code for arrays
public static void Recurse(int newArray,int i)
{
for(int j=i;j<newArray.length;j++)
{
Swap(newArray,i,j);
if(i<(newArray.length-1))
{
Recurse(newArray,i+1);
}
if(i==(newArray.length-1))
{
Display(newArray);
}
}
}
public static void Display(int Array)
{
for(int i: Array)
{
System.out.print(i);
}
System.out.println();
}
public static void main(String args)
{
Recurse(new int {1,2,3},0);
}
output
123
132
312
321
123
132
Does not work correctly,So i was wondering what went wrong here.
java arrays string permutation
asked Nov 20 '18 at 17:18
user9169047user9169047
183
1
Could you post the code for theSwapmethod?
– GBlodgett
Nov 20 '18 at 17:30
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
Are you using the sameSwapmethod for yourArrayversion?
– GBlodgett
Nov 20 '18 at 17:36
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52
|
show 1 more comment
I was trying to permutate using java and i noticed my code works for strings but the same algorithm does not work correctly on Arrays
here's the code for strings
public static void Recurse(String s,int i)
{
for(int j=i;j<(s.length());j++)
{
s=Swap(s,i,j);
if(i<(s.length()-1))
{
Recurse(s,i+1);
}
if(i==(s.length()-1))
{
Display(s);
}
}
}
public static String Swap(String newString,int i,int j)
{
char newChar=newString.toCharArray();
char temp=newChar[i];
newChar[i]=newChar[j];
newChar[j]=temp;
return String.valueOf(newChar);
}
public static void Display(String s)
{
System.out.println(s);
}
public static void main(String args)
{
String i=new String("123");
Recurse(i,0);
}
output:
123
132
213
231
312
321
works fine.
then here is the code for arrays
public static void Recurse(int newArray,int i)
{
for(int j=i;j<newArray.length;j++)
{
Swap(newArray,i,j);
if(i<(newArray.length-1))
{
Recurse(newArray,i+1);
}
if(i==(newArray.length-1))
{
Display(newArray);
}
}
}
public static void Display(int Array)
{
for(int i: Array)
{
System.out.print(i);
}
System.out.println();
}
public static void main(String args)
{
Recurse(new int {1,2,3},0);
}
output
123
132
312
321
123
132
Does not work correctly,So i was wondering what went wrong here.
java arrays string permutation
asked Nov 20 '18 at 17:18
user9169047user9169047
183
I was trying to permutate using java and i noticed my code works for strings but the same algorithm does not work correctly on Arrays
here's the code for strings
public static void Recurse(String s,int i)
{
for(int j=i;j<(s.length());j++)
{
s=Swap(s,i,j);
if(i<(s.length()-1))
{
Recurse(s,i+1);
}
if(i==(s.length()-1))
{
Display(s);
}
}
}
public static String Swap(String newString,int i,int j)
{
char newChar=newString.toCharArray();
char temp=newChar[i];
newChar[i]=newChar[j];
newChar[j]=temp;
return String.valueOf(newChar);
}
public static void Display(String s)
{
System.out.println(s);
}
public static void main(String args)
{
String i=new String("123");
Recurse(i,0);
}
output:
123
132
213
231
312
321
works fine.
then here is the code for arrays
public static void Recurse(int newArray,int i)
{
for(int j=i;j<newArray.length;j++)
{
Swap(newArray,i,j);
if(i<(newArray.length-1))
{
Recurse(newArray,i+1);
}
if(i==(newArray.length-1))
{
Display(newArray);
}
}
}
public static void Display(int Array)
{
for(int i: Array)
{
System.out.print(i);
}
System.out.println();
}
public static void main(String args)
{
Recurse(new int {1,2,3},0);
}
output
123
132
312
321
123
132
Does not work correctly,So i was wondering what went wrong here.
java arrays string permutation
java arrays string permutation
asked Nov 20 '18 at 17:18
user9169047user9169047
183
asked Nov 20 '18 at 17:18
user9169047user9169047
183
edited Nov 20 '18 at 17:37
user9169047
asked Nov 20 '18 at 17:18
user9169047user9169047
183
asked Nov 20 '18 at 17:18
user9169047user9169047
183
asked Nov 20 '18 at 17:18
user9169047user9169047
183
183
1
Could you post the code for theSwapmethod?
– GBlodgett
Nov 20 '18 at 17:30
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
Are you using the sameSwapmethod for yourArrayversion?
– GBlodgett
Nov 20 '18 at 17:36
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52
|
show 1 more comment
1
Could you post the code for theSwapmethod?
– GBlodgett
Nov 20 '18 at 17:30
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
Are you using the sameSwapmethod for yourArrayversion?
– GBlodgett
Nov 20 '18 at 17:36
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52
1
1
Could you post the code for the
Swap method?– GBlodgett
Nov 20 '18 at 17:30
Could you post the code for the
Swap method?– GBlodgett
Nov 20 '18 at 17:30
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
Are you using the same
Swap method for your Array version?– GBlodgett
Nov 20 '18 at 17:36
Are you using the same
Swap method for your Array version?– GBlodgett
Nov 20 '18 at 17:36
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52
|
show 1 more comment
2 Answers
2
active
oldest
votes
The output seems to be correct with below code:
public static void recurse(int newArray, int i) {
for (int j = i; j < newArray.length; j++) {
newArray = swap(newArray, i, j);
if (i < (newArray.length - 1)) {
recurse(newArray, i + 1);
}
if (i == (newArray.length - 1)) {
display(newArray);
}
}
}
private static int swap(int a, int i, int j) {
int newArray = a.clone();
int temp = newArray[i];
newArray[i] = newArray[j];
newArray[j] = temp;
return newArray;
}
public static void display(int Array) {
for (int i : Array) {
System.out.print(i);
}
System.out.println();
}
public static void main(String args) {
recurse(new int{1, 2, 3}, 0);
}
the output is :
123
132
213
231
312
321
I just made few changes in swap methods code, I made it similar to the one which you used in Strings swap method.
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
add a comment |
Strings are immutable, so when Recurse(String s,int i) calls Recurse(s,i+1);, the value of s remains unchanged.
Arrays are mutable, so when Recurse(int newArray,int i) calls Recurse(newArray,i+1);, the value of newArray is changed.
That is why they behave differently.
Note: When I say "the value of" I'm referring to the value of the object, not the value of the reference variable itself.
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The output seems to be correct with below code:
public static void recurse(int newArray, int i) {
for (int j = i; j < newArray.length; j++) {
newArray = swap(newArray, i, j);
if (i < (newArray.length - 1)) {
recurse(newArray, i + 1);
}
if (i == (newArray.length - 1)) {
display(newArray);
}
}
}
private static int swap(int a, int i, int j) {
int newArray = a.clone();
int temp = newArray[i];
newArray[i] = newArray[j];
newArray[j] = temp;
return newArray;
}
public static void display(int Array) {
for (int i : Array) {
System.out.print(i);
}
System.out.println();
}
public static void main(String args) {
recurse(new int{1, 2, 3}, 0);
}
the output is :
123
132
213
231
312
321
I just made few changes in swap methods code, I made it similar to the one which you used in Strings swap method.
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
add a comment |
The output seems to be correct with below code:
public static void recurse(int newArray, int i) {
for (int j = i; j < newArray.length; j++) {
newArray = swap(newArray, i, j);
if (i < (newArray.length - 1)) {
recurse(newArray, i + 1);
}
if (i == (newArray.length - 1)) {
display(newArray);
}
}
}
private static int swap(int a, int i, int j) {
int newArray = a.clone();
int temp = newArray[i];
newArray[i] = newArray[j];
newArray[j] = temp;
return newArray;
}
public static void display(int Array) {
for (int i : Array) {
System.out.print(i);
}
System.out.println();
}
public static void main(String args) {
recurse(new int{1, 2, 3}, 0);
}
the output is :
123
132
213
231
312
321
I just made few changes in swap methods code, I made it similar to the one which you used in Strings swap method.
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
add a comment |
The output seems to be correct with below code:
public static void recurse(int newArray, int i) {
for (int j = i; j < newArray.length; j++) {
newArray = swap(newArray, i, j);
if (i < (newArray.length - 1)) {
recurse(newArray, i + 1);
}
if (i == (newArray.length - 1)) {
display(newArray);
}
}
}
private static int swap(int a, int i, int j) {
int newArray = a.clone();
int temp = newArray[i];
newArray[i] = newArray[j];
newArray[j] = temp;
return newArray;
}
public static void display(int Array) {
for (int i : Array) {
System.out.print(i);
}
System.out.println();
}
public static void main(String args) {
recurse(new int{1, 2, 3}, 0);
}
the output is :
123
132
213
231
312
321
I just made few changes in swap methods code, I made it similar to the one which you used in Strings swap method.
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
The output seems to be correct with below code:
public static void recurse(int newArray, int i) {
for (int j = i; j < newArray.length; j++) {
newArray = swap(newArray, i, j);
if (i < (newArray.length - 1)) {
recurse(newArray, i + 1);
}
if (i == (newArray.length - 1)) {
display(newArray);
}
}
}
private static int swap(int a, int i, int j) {
int newArray = a.clone();
int temp = newArray[i];
newArray[i] = newArray[j];
newArray[j] = temp;
return newArray;
}
public static void display(int Array) {
for (int i : Array) {
System.out.print(i);
}
System.out.println();
}
public static void main(String args) {
recurse(new int{1, 2, 3}, 0);
}
the output is :
123
132
213
231
312
321
I just made few changes in swap methods code, I made it similar to the one which you used in Strings swap method.
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
answered Nov 20 '18 at 17:43
eatSleepCodeeatSleepCode
2,26132463
2,26132463
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
add a comment |
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
Thanks ,this works ,but what exactly is the difference between this swap function and the other one that is making them produce different output
– user9169047
Nov 20 '18 at 18:27
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
@user9169047 Array is a referenceType, on passing it to the method the actual values in it will be changed. and the logic to permute numbers needs the array to be in its earlier state.
– eatSleepCode
Nov 20 '18 at 18:31
add a comment |
Strings are immutable, so when Recurse(String s,int i) calls Recurse(s,i+1);, the value of s remains unchanged.
Arrays are mutable, so when Recurse(int newArray,int i) calls Recurse(newArray,i+1);, the value of newArray is changed.
That is why they behave differently.
Note: When I say "the value of" I'm referring to the value of the object, not the value of the reference variable itself.
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
add a comment |
Strings are immutable, so when Recurse(String s,int i) calls Recurse(s,i+1);, the value of s remains unchanged.
Arrays are mutable, so when Recurse(int newArray,int i) calls Recurse(newArray,i+1);, the value of newArray is changed.
That is why they behave differently.
Note: When I say "the value of" I'm referring to the value of the object, not the value of the reference variable itself.
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
add a comment |
Strings are immutable, so when Recurse(String s,int i) calls Recurse(s,i+1);, the value of s remains unchanged.
Arrays are mutable, so when Recurse(int newArray,int i) calls Recurse(newArray,i+1);, the value of newArray is changed.
That is why they behave differently.
Note: When I say "the value of" I'm referring to the value of the object, not the value of the reference variable itself.
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
Strings are immutable, so when Recurse(String s,int i) calls Recurse(s,i+1);, the value of s remains unchanged.
Arrays are mutable, so when Recurse(int newArray,int i) calls Recurse(newArray,i+1);, the value of newArray is changed.
That is why they behave differently.
Note: When I say "the value of" I'm referring to the value of the object, not the value of the reference variable itself.
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
answered Nov 20 '18 at 17:54
AndreasAndreas
76.3k463123
76.3k463123
add a comment |
add a comment |
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Could you post the code for the
Swapmethod?– GBlodgett
Nov 20 '18 at 17:30
i have reedited the code
– user9169047
Nov 20 '18 at 17:33
Are you using the same
Swapmethod for yourArrayversion?– GBlodgett
Nov 20 '18 at 17:36
no,the array was swapped using temp variables,but strings are immutable so i had to convert to charArray then return a new String object to swap the string.
– user9169047
Nov 20 '18 at 17:41
You should avoid the String -> char -> String conversion. Just operate on char and to the from/to String conversion and the beginning/end.
– Robert
Nov 20 '18 at 17:52