Insertion/Deletion of Doubly Linked List in constant time O(1)
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a) I have seen a lot of examples of this question in stackoverflow but I am still having problems understanding how does it know which node it is referring when call functions like insertAfter(Node n, Object o). If we say insert after node 2, how does linkedlist know which node is node 2?
b) in the previous posts in stackoverflow, it is said there is a pointer to a node that you want to insert after or before, that is why we get constant time operations. does that mean, just like we have to head and tail in linked list, we also have a pointer to each node?
would really appreciate help in understanding this.
list constants
edited Nov 11 at 7:18
quant
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
add a comment |
up vote
0
down vote
favorite
a) I have seen a lot of examples of this question in stackoverflow but I am still having problems understanding how does it know which node it is referring when call functions like insertAfter(Node n, Object o). If we say insert after node 2, how does linkedlist know which node is node 2?
b) in the previous posts in stackoverflow, it is said there is a pointer to a node that you want to insert after or before, that is why we get constant time operations. does that mean, just like we have to head and tail in linked list, we also have a pointer to each node?
would really appreciate help in understanding this.
list constants
edited Nov 11 at 7:18
quant
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
a) I have seen a lot of examples of this question in stackoverflow but I am still having problems understanding how does it know which node it is referring when call functions like insertAfter(Node n, Object o). If we say insert after node 2, how does linkedlist know which node is node 2?
b) in the previous posts in stackoverflow, it is said there is a pointer to a node that you want to insert after or before, that is why we get constant time operations. does that mean, just like we have to head and tail in linked list, we also have a pointer to each node?
would really appreciate help in understanding this.
list constants
edited Nov 11 at 7:18
quant
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
a) I have seen a lot of examples of this question in stackoverflow but I am still having problems understanding how does it know which node it is referring when call functions like insertAfter(Node n, Object o). If we say insert after node 2, how does linkedlist know which node is node 2?
b) in the previous posts in stackoverflow, it is said there is a pointer to a node that you want to insert after or before, that is why we get constant time operations. does that mean, just like we have to head and tail in linked list, we also have a pointer to each node?
would really appreciate help in understanding this.
list constants
list constants
edited Nov 11 at 7:18
quant
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
edited Nov 11 at 7:18
quant
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
edited Nov 11 at 7:18
quant
1,53911526
edited Nov 11 at 7:18
quant
1,53911526
edited Nov 11 at 7:18
quant
1,53911526
1,53911526
asked Sep 13 '12 at 5:47
ueg1990
3491930
asked Sep 13 '12 at 5:47
ueg1990
3491930
asked Sep 13 '12 at 5:47
ueg1990
3491930
3491930
add a comment |
add a comment |
1 Answer
1
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votes
up vote
2
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If deletion is done using key then your point is valid because we dont know the position of the element to be deleted and thus finding key in the list makes its time linear in the length of the queue. But at places where it is written to be Constant time is deletion by address. So one may always move to that address and delete the node in constant time.
Note : This ain't possible with singly linked list.
answered Nov 27 '12 at 10:31
timedout
168210
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If deletion is done using key then your point is valid because we dont know the position of the element to be deleted and thus finding key in the list makes its time linear in the length of the queue. But at places where it is written to be Constant time is deletion by address. So one may always move to that address and delete the node in constant time.
Note : This ain't possible with singly linked list.
answered Nov 27 '12 at 10:31
timedout
168210
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
add a comment |
up vote
2
down vote
If deletion is done using key then your point is valid because we dont know the position of the element to be deleted and thus finding key in the list makes its time linear in the length of the queue. But at places where it is written to be Constant time is deletion by address. So one may always move to that address and delete the node in constant time.
Note : This ain't possible with singly linked list.
answered Nov 27 '12 at 10:31
timedout
168210
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
add a comment |
up vote
2
down vote
up vote
2
down vote
If deletion is done using key then your point is valid because we dont know the position of the element to be deleted and thus finding key in the list makes its time linear in the length of the queue. But at places where it is written to be Constant time is deletion by address. So one may always move to that address and delete the node in constant time.
Note : This ain't possible with singly linked list.
answered Nov 27 '12 at 10:31
timedout
168210
If deletion is done using key then your point is valid because we dont know the position of the element to be deleted and thus finding key in the list makes its time linear in the length of the queue. But at places where it is written to be Constant time is deletion by address. So one may always move to that address and delete the node in constant time.
Note : This ain't possible with singly linked list.
answered Nov 27 '12 at 10:31
timedout
168210
answered Nov 27 '12 at 10:31
timedout
168210
answered Nov 27 '12 at 10:31
timedout
168210
answered Nov 27 '12 at 10:31
timedout
168210
168210
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
add a comment |
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
why is it not possible?
– dove
Nov 27 '12 at 10:59
why is it not possible?
– dove
Nov 27 '12 at 10:59
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
because in case of singly linked list you don't have location of parent to connect to next.
– timedout
Nov 28 '12 at 4:30
add a comment |
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