Write hashfunction in Java
I am trying to create the data structure of a HashMap() in Java. The HashMap has to work for a maximum of N = 1000 operations and the keys are only positive integers. What I did is the following:
class MyHashMap {
final ListNode nodes = new ListNode[1000];
// "put", "get" and "remove" methods which take
// collisions into account using "chaining".
}
to decide the placement of my new key - value pair in nodes I always need to compute an index. I use the function:
public int getIndex(int key) { return Integer.hashCode(key) % nodes.length;}
which returns an integer between 0 and nodes.length. But how can I write a Hashfunction in Java on my own that maps integers to some index without using Integer.hashMap(key)?
Also the procedure is fine, I don't really need a code.
java hashmap hash-function
asked Nov 12 at 14:26
Giacomo
886
add a comment |
I am trying to create the data structure of a HashMap() in Java. The HashMap has to work for a maximum of N = 1000 operations and the keys are only positive integers. What I did is the following:
class MyHashMap {
final ListNode nodes = new ListNode[1000];
// "put", "get" and "remove" methods which take
// collisions into account using "chaining".
}
to decide the placement of my new key - value pair in nodes I always need to compute an index. I use the function:
public int getIndex(int key) { return Integer.hashCode(key) % nodes.length;}
which returns an integer between 0 and nodes.length. But how can I write a Hashfunction in Java on my own that maps integers to some index without using Integer.hashMap(key)?
Also the procedure is fine, I don't really need a code.
java hashmap hash-function
asked Nov 12 at 14:26
Giacomo
886
add a comment |
I am trying to create the data structure of a HashMap() in Java. The HashMap has to work for a maximum of N = 1000 operations and the keys are only positive integers. What I did is the following:
class MyHashMap {
final ListNode nodes = new ListNode[1000];
// "put", "get" and "remove" methods which take
// collisions into account using "chaining".
}
to decide the placement of my new key - value pair in nodes I always need to compute an index. I use the function:
public int getIndex(int key) { return Integer.hashCode(key) % nodes.length;}
which returns an integer between 0 and nodes.length. But how can I write a Hashfunction in Java on my own that maps integers to some index without using Integer.hashMap(key)?
Also the procedure is fine, I don't really need a code.
java hashmap hash-function
asked Nov 12 at 14:26
Giacomo
886
I am trying to create the data structure of a HashMap() in Java. The HashMap has to work for a maximum of N = 1000 operations and the keys are only positive integers. What I did is the following:
class MyHashMap {
final ListNode nodes = new ListNode[1000];
// "put", "get" and "remove" methods which take
// collisions into account using "chaining".
}
to decide the placement of my new key - value pair in nodes I always need to compute an index. I use the function:
public int getIndex(int key) { return Integer.hashCode(key) % nodes.length;}
which returns an integer between 0 and nodes.length. But how can I write a Hashfunction in Java on my own that maps integers to some index without using Integer.hashMap(key)?
Also the procedure is fine, I don't really need a code.
java hashmap hash-function
java hashmap hash-function
asked Nov 12 at 14:26
Giacomo
886
asked Nov 12 at 14:26
Giacomo
886
edited Nov 12 at 14:48
asked Nov 12 at 14:26
Giacomo
886
asked Nov 12 at 14:26
Giacomo
886
asked Nov 12 at 14:26
Giacomo
886
886
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
A simple strategy for hashing an int value is to multiply by a large constant. If you don't use a constant, you can get very poor collision rates depending on your key distribution. It's is still possible to get poor key distribution, however it is less likely for real data.
NOTE: Unless you know the key cannot be negative, you should use Math.abs to ensure it is non-negative.
static final int K = 0x7A646E4D; // some large prime.
public int getIndex(int key) {
return Math.abs(key * K % nodes.length);
}
A faster solution is to drop the use of % use a multiplication and shift. e.g.
public int getIndex(int key) {
return (int) ((key * K & 0xFFFF_FFFFL) * nodes.length >>> 32);
}
What this does is turn key * K into a fraction and is faster than using %
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value likeintor an array or values likecharwhich will require a loop but the principles are the same.
– Peter Lawrey
Nov 12 at 17:10
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
add a comment |
Integer.hashCode(key) just returns key as result. You could write:
public int getIndex(int key) {
return key % nodes.length;
}
This way, you don't use Integer.hashCode(key).
Without seeing your class ListNode, you could have a problem with your MyHashMap structure since you are setting many elements to the same array element. For instance the key 3 and 1003 will be saved in the same node element. You must design MyHashMap to be a list of list (or an array of list or something similar).
answered Nov 12 at 14:38
LaurentG
7,87573553
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:-1 % nodes.length == -1
– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
add a comment |
Well, first of all the hash is something that is unique or a very rarely repeating value.
Assuming that your values will be evenly distributed you can use the remainder of division as a hash value for your key.
public int get(int key) {
return (-1 ^ key) % nodes.length;
}
answered Nov 12 at 14:40
ETO
1,589320
Note: ifkeyis negative, this can return a negative number.
– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
A simple strategy for hashing an int value is to multiply by a large constant. If you don't use a constant, you can get very poor collision rates depending on your key distribution. It's is still possible to get poor key distribution, however it is less likely for real data.
NOTE: Unless you know the key cannot be negative, you should use Math.abs to ensure it is non-negative.
static final int K = 0x7A646E4D; // some large prime.
public int getIndex(int key) {
return Math.abs(key * K % nodes.length);
}
A faster solution is to drop the use of % use a multiplication and shift. e.g.
public int getIndex(int key) {
return (int) ((key * K & 0xFFFF_FFFFL) * nodes.length >>> 32);
}
What this does is turn key * K into a fraction and is faster than using %
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value likeintor an array or values likecharwhich will require a loop but the principles are the same.
– Peter Lawrey
Nov 12 at 17:10
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
add a comment |
A simple strategy for hashing an int value is to multiply by a large constant. If you don't use a constant, you can get very poor collision rates depending on your key distribution. It's is still possible to get poor key distribution, however it is less likely for real data.
NOTE: Unless you know the key cannot be negative, you should use Math.abs to ensure it is non-negative.
static final int K = 0x7A646E4D; // some large prime.
public int getIndex(int key) {
return Math.abs(key * K % nodes.length);
}
A faster solution is to drop the use of % use a multiplication and shift. e.g.
public int getIndex(int key) {
return (int) ((key * K & 0xFFFF_FFFFL) * nodes.length >>> 32);
}
What this does is turn key * K into a fraction and is faster than using %
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value likeintor an array or values likecharwhich will require a loop but the principles are the same.
– Peter Lawrey
Nov 12 at 17:10
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
add a comment |
A simple strategy for hashing an int value is to multiply by a large constant. If you don't use a constant, you can get very poor collision rates depending on your key distribution. It's is still possible to get poor key distribution, however it is less likely for real data.
NOTE: Unless you know the key cannot be negative, you should use Math.abs to ensure it is non-negative.
static final int K = 0x7A646E4D; // some large prime.
public int getIndex(int key) {
return Math.abs(key * K % nodes.length);
}
A faster solution is to drop the use of % use a multiplication and shift. e.g.
public int getIndex(int key) {
return (int) ((key * K & 0xFFFF_FFFFL) * nodes.length >>> 32);
}
What this does is turn key * K into a fraction and is faster than using %
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
A simple strategy for hashing an int value is to multiply by a large constant. If you don't use a constant, you can get very poor collision rates depending on your key distribution. It's is still possible to get poor key distribution, however it is less likely for real data.
NOTE: Unless you know the key cannot be negative, you should use Math.abs to ensure it is non-negative.
static final int K = 0x7A646E4D; // some large prime.
public int getIndex(int key) {
return Math.abs(key * K % nodes.length);
}
A faster solution is to drop the use of % use a multiplication and shift. e.g.
public int getIndex(int key) {
return (int) ((key * K & 0xFFFF_FFFFL) * nodes.length >>> 32);
}
What this does is turn key * K into a fraction and is faster than using %
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
edited Nov 12 at 14:46
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
answered Nov 12 at 14:41
Peter Lawrey
440k55558957
440k55558957
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value likeintor an array or values likecharwhich will require a loop but the principles are the same.
– Peter Lawrey
Nov 12 at 17:10
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
add a comment |
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value likeintor an array or values likecharwhich will require a loop but the principles are the same.
– Peter Lawrey
Nov 12 at 17:10
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
1
1
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
Seems like you first take the lowest 32 bits after multiplication, and in the end you use only the upper 32 bits?
– Marko Topolnik
Nov 12 at 14:57
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
do you maybe know where these things are explained ? e.g. how to hash an integer, a string ecc.
– Giacomo
Nov 12 at 15:02
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@MarkoTopolnik In the second example, I start with the lower bits of a 64-bit value which after shifting by 32-bits at the end, act like a fraction.
– Peter Lawrey
Nov 12 at 17:09
@Giacomo all data are numbers. You either have a single value like
int or an array or values like char which will require a loop but the principles are the same.– Peter Lawrey
Nov 12 at 17:10
@Giacomo all data are numbers. You either have a single value like
int or an array or values like char which will require a loop but the principles are the same.– Peter Lawrey
Nov 12 at 17:10
1
1
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
I see now. You use the 32-bit number as a scale factor which is effectively in the range (0,1).
– Marko Topolnik
Nov 12 at 18:04
add a comment |
Integer.hashCode(key) just returns key as result. You could write:
public int getIndex(int key) {
return key % nodes.length;
}
This way, you don't use Integer.hashCode(key).
Without seeing your class ListNode, you could have a problem with your MyHashMap structure since you are setting many elements to the same array element. For instance the key 3 and 1003 will be saved in the same node element. You must design MyHashMap to be a list of list (or an array of list or something similar).
answered Nov 12 at 14:38
LaurentG
7,87573553
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:-1 % nodes.length == -1
– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
add a comment |
Integer.hashCode(key) just returns key as result. You could write:
public int getIndex(int key) {
return key % nodes.length;
}
This way, you don't use Integer.hashCode(key).
Without seeing your class ListNode, you could have a problem with your MyHashMap structure since you are setting many elements to the same array element. For instance the key 3 and 1003 will be saved in the same node element. You must design MyHashMap to be a list of list (or an array of list or something similar).
answered Nov 12 at 14:38
LaurentG
7,87573553
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:-1 % nodes.length == -1
– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
add a comment |
Integer.hashCode(key) just returns key as result. You could write:
public int getIndex(int key) {
return key % nodes.length;
}
This way, you don't use Integer.hashCode(key).
Without seeing your class ListNode, you could have a problem with your MyHashMap structure since you are setting many elements to the same array element. For instance the key 3 and 1003 will be saved in the same node element. You must design MyHashMap to be a list of list (or an array of list or something similar).
answered Nov 12 at 14:38
LaurentG
7,87573553
Integer.hashCode(key) just returns key as result. You could write:
public int getIndex(int key) {
return key % nodes.length;
}
This way, you don't use Integer.hashCode(key).
Without seeing your class ListNode, you could have a problem with your MyHashMap structure since you are setting many elements to the same array element. For instance the key 3 and 1003 will be saved in the same node element. You must design MyHashMap to be a list of list (or an array of list or something similar).
answered Nov 12 at 14:38
LaurentG
7,87573553
answered Nov 12 at 14:38
LaurentG
7,87573553
answered Nov 12 at 14:38
LaurentG
7,87573553
answered Nov 12 at 14:38
LaurentG
7,87573553
7,87573553
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:-1 % nodes.length == -1
– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
add a comment |
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:-1 % nodes.length == -1
– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Maybe my question what not so clear. Doing it like I did, Integer.hashCode(key) will return an Integer value which can be everything. Taking % nodes.length this will ensure that this value has now a number which is in the range of my array nodes. If I get a value which corresponds to a cell which is already occupied, I will create (in my methods, this can not be seen in the question) a link to a new node to prevent the collision. So I don't really have the problem you are mentioning.
– Giacomo
Nov 12 at 14:43
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
Well you must still consider the collisions since they are possible. Of course there is hash function which reduces the collisions, but it was not in your question.
– LaurentG
Nov 12 at 14:46
@Giacomo Note:
-1 % nodes.length == -1– Peter Lawrey
Nov 12 at 14:47
@Giacomo Note:
-1 % nodes.length == -1– Peter Lawrey
Nov 12 at 14:47
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
@PeterLawrey i did not mention that keys are potive integers, need to edit it
– Giacomo
Nov 12 at 14:48
add a comment |
Well, first of all the hash is something that is unique or a very rarely repeating value.
Assuming that your values will be evenly distributed you can use the remainder of division as a hash value for your key.
public int get(int key) {
return (-1 ^ key) % nodes.length;
}
answered Nov 12 at 14:40
ETO
1,589320
Note: ifkeyis negative, this can return a negative number.
– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
add a comment |
Well, first of all the hash is something that is unique or a very rarely repeating value.
Assuming that your values will be evenly distributed you can use the remainder of division as a hash value for your key.
public int get(int key) {
return (-1 ^ key) % nodes.length;
}
answered Nov 12 at 14:40
ETO
1,589320
Note: ifkeyis negative, this can return a negative number.
– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
add a comment |
Well, first of all the hash is something that is unique or a very rarely repeating value.
Assuming that your values will be evenly distributed you can use the remainder of division as a hash value for your key.
public int get(int key) {
return (-1 ^ key) % nodes.length;
}
answered Nov 12 at 14:40
ETO
1,589320
Well, first of all the hash is something that is unique or a very rarely repeating value.
Assuming that your values will be evenly distributed you can use the remainder of division as a hash value for your key.
public int get(int key) {
return (-1 ^ key) % nodes.length;
}
answered Nov 12 at 14:40
ETO
1,589320
edited Nov 12 at 14:58
answered Nov 12 at 14:40
ETO
1,589320
answered Nov 12 at 14:40
ETO
1,589320
answered Nov 12 at 14:40
ETO
1,589320
1,589320
Note: ifkeyis negative, this can return a negative number.
– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
add a comment |
Note: ifkeyis negative, this can return a negative number.
– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
Note: if
key is negative, this can return a negative number.– Peter Lawrey
Nov 12 at 14:48
Note: if
key is negative, this can return a negative number.– Peter Lawrey
Nov 12 at 14:48
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
@PeterLawrey Yes, you're right. Fixed the answer.
– ETO
Nov 12 at 14:59
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