Clean sequences from a LinkedList











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Edited
I have to remove from a LinkedList sequences which their sum equals to 5.



for example for this Linked list: {3,2, 1, 4, 10,3,3,2,1} I get {10, 3, 3, 2, 1}. I dont really know why it's like that, it should remove 3,2 also but it dosent.



void LinkedList::cleanFive(LinkedList& list)
{
head = list.GetHead();

Node* end = head;
Node* headd = head;
Node* flag = head;
Node* curr = head;

int sum = 0;


while (end) {
sum += end->data;

if (sum < 5) {
end = end->next;
}
if (sum == 5) {
end = end->next;
headd = end;
flag = headd;
head = flag;
sum = 0;
}
if (sum > 5) {
flag = head;
curr = end;
end = end->next;
sum = 0;
}
}


}










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  • 1




    Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
    – molbdnilo
    Nov 22 at 14:56










  • The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
    – molbdnilo
    Nov 22 at 15:00










  • In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
    – Igor Tandetnik
    Nov 22 at 15:09








  • 1




    What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
    – Bob__
    Nov 22 at 15:16










  • @Bob__ in this particular case: { }
    – Benjamin Yakobi
    Nov 22 at 16:15















up vote
-1
down vote

favorite












Edited
I have to remove from a LinkedList sequences which their sum equals to 5.



for example for this Linked list: {3,2, 1, 4, 10,3,3,2,1} I get {10, 3, 3, 2, 1}. I dont really know why it's like that, it should remove 3,2 also but it dosent.



void LinkedList::cleanFive(LinkedList& list)
{
head = list.GetHead();

Node* end = head;
Node* headd = head;
Node* flag = head;
Node* curr = head;

int sum = 0;


while (end) {
sum += end->data;

if (sum < 5) {
end = end->next;
}
if (sum == 5) {
end = end->next;
headd = end;
flag = headd;
head = flag;
sum = 0;
}
if (sum > 5) {
flag = head;
curr = end;
end = end->next;
sum = 0;
}
}


}










share|improve this question




















  • 1




    Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
    – molbdnilo
    Nov 22 at 14:56










  • The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
    – molbdnilo
    Nov 22 at 15:00










  • In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
    – Igor Tandetnik
    Nov 22 at 15:09








  • 1




    What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
    – Bob__
    Nov 22 at 15:16










  • @Bob__ in this particular case: { }
    – Benjamin Yakobi
    Nov 22 at 16:15













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Edited
I have to remove from a LinkedList sequences which their sum equals to 5.



for example for this Linked list: {3,2, 1, 4, 10,3,3,2,1} I get {10, 3, 3, 2, 1}. I dont really know why it's like that, it should remove 3,2 also but it dosent.



void LinkedList::cleanFive(LinkedList& list)
{
head = list.GetHead();

Node* end = head;
Node* headd = head;
Node* flag = head;
Node* curr = head;

int sum = 0;


while (end) {
sum += end->data;

if (sum < 5) {
end = end->next;
}
if (sum == 5) {
end = end->next;
headd = end;
flag = headd;
head = flag;
sum = 0;
}
if (sum > 5) {
flag = head;
curr = end;
end = end->next;
sum = 0;
}
}


}










share|improve this question















Edited
I have to remove from a LinkedList sequences which their sum equals to 5.



for example for this Linked list: {3,2, 1, 4, 10,3,3,2,1} I get {10, 3, 3, 2, 1}. I dont really know why it's like that, it should remove 3,2 also but it dosent.



void LinkedList::cleanFive(LinkedList& list)
{
head = list.GetHead();

Node* end = head;
Node* headd = head;
Node* flag = head;
Node* curr = head;

int sum = 0;


while (end) {
sum += end->data;

if (sum < 5) {
end = end->next;
}
if (sum == 5) {
end = end->next;
headd = end;
flag = headd;
head = flag;
sum = 0;
}
if (sum > 5) {
flag = head;
curr = end;
end = end->next;
sum = 0;
}
}


}







c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 at 17:10

























asked Nov 22 at 14:48









Benjamin Yakobi

525




525








  • 1




    Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
    – molbdnilo
    Nov 22 at 14:56










  • The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
    – molbdnilo
    Nov 22 at 15:00










  • In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
    – Igor Tandetnik
    Nov 22 at 15:09








  • 1




    What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
    – Bob__
    Nov 22 at 15:16










  • @Bob__ in this particular case: { }
    – Benjamin Yakobi
    Nov 22 at 16:15














  • 1




    Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
    – molbdnilo
    Nov 22 at 14:56










  • The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
    – molbdnilo
    Nov 22 at 15:00










  • In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
    – Igor Tandetnik
    Nov 22 at 15:09








  • 1




    What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
    – Bob__
    Nov 22 at 15:16










  • @Bob__ in this particular case: { }
    – Benjamin Yakobi
    Nov 22 at 16:15








1




1




Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
– molbdnilo
Nov 22 at 14:56




Why does this take a parameter and why does it take over the parameter's head node? Shouldn't it remove such a sequence from itself?
– molbdnilo
Nov 22 at 14:56












The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
– molbdnilo
Nov 22 at 15:00




The best tools for fixing (and creating) pointer-related code is pen(cil) and paper. Draw your list and figure out what should happen. Then compare that to what does happen.
– molbdnilo
Nov 22 at 15:00












In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
– Igor Tandetnik
Nov 22 at 15:09






In your example, you try 3 + 4, it exceeds 5, whereupon you start counting again from 1 and miss 4 + 1 sequence. You need to advance curr by one element, not all the way to the end; and at the same time, subtract curr->data from sum to account for the sequence getting shorter.
– Igor Tandetnik
Nov 22 at 15:09






1




1




What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
– Bob__
Nov 22 at 15:16




What is the expected output in case of {3, 4, 3, 2, 1, 2, 5}, {}?
– Bob__
Nov 22 at 15:16












@Bob__ in this particular case: { }
– Benjamin Yakobi
Nov 22 at 16:15




@Bob__ in this particular case: { }
– Benjamin Yakobi
Nov 22 at 16:15












1 Answer
1






active

oldest

votes

















up vote
1
down vote













curr = head;
end = head;
currprev = nullptr;

std::vector<Node*> flag;
while(end) {
sum += end->data;
if (sum < 5) {
flag.push_back(end);
end = end->next;
}

if (sum > 5) {
currprev = curr;
curr = curr->next;
end = curr;
sum = 0;
empty_flag()
}

if (sum == 5) {
curr = end->next;
end = curr;
sum = 0;
delete_nodes_from_flag_from_linked_list()
empty_flag()
}


}



Here is a possible solution. flag is in this case a vector of nodes which are saved until the sum is proven to be 5 or greater as 5. With flag there are two operations which need to be made: empy_flag() - deletes all entries from the vector and delete_nodes_from_flag_from_linked_list() - that deletes the elements from flag from the linked list.



Code for delete_nodes_from_flag_from_linked_list() should be as follows:



auto it = flag.end();
it--;
Node* last = *it;

if (currprev)
currprev->next = last->next;
else
head = last->next

for (auto f : flag)
delete f;


where currprev is the element just before curr. You have to keep track of this element from the beginning, unless your linked list elements keep pointers to previous elements. I am updating the code from above. head is the start of the list.






share|improve this answer























  • how do I implement this? delete_nodes_from_flag_from_linked_list()
    – Benjamin Yakobi
    Nov 22 at 16:27








  • 1




    Please see revised anwer.
    – Cristi
    Nov 22 at 16:53










  • Could you please mark as anwer if it is a correct answer ?
    – Cristi
    Nov 23 at 7:30











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1 Answer
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up vote
1
down vote













curr = head;
end = head;
currprev = nullptr;

std::vector<Node*> flag;
while(end) {
sum += end->data;
if (sum < 5) {
flag.push_back(end);
end = end->next;
}

if (sum > 5) {
currprev = curr;
curr = curr->next;
end = curr;
sum = 0;
empty_flag()
}

if (sum == 5) {
curr = end->next;
end = curr;
sum = 0;
delete_nodes_from_flag_from_linked_list()
empty_flag()
}


}



Here is a possible solution. flag is in this case a vector of nodes which are saved until the sum is proven to be 5 or greater as 5. With flag there are two operations which need to be made: empy_flag() - deletes all entries from the vector and delete_nodes_from_flag_from_linked_list() - that deletes the elements from flag from the linked list.



Code for delete_nodes_from_flag_from_linked_list() should be as follows:



auto it = flag.end();
it--;
Node* last = *it;

if (currprev)
currprev->next = last->next;
else
head = last->next

for (auto f : flag)
delete f;


where currprev is the element just before curr. You have to keep track of this element from the beginning, unless your linked list elements keep pointers to previous elements. I am updating the code from above. head is the start of the list.






share|improve this answer























  • how do I implement this? delete_nodes_from_flag_from_linked_list()
    – Benjamin Yakobi
    Nov 22 at 16:27








  • 1




    Please see revised anwer.
    – Cristi
    Nov 22 at 16:53










  • Could you please mark as anwer if it is a correct answer ?
    – Cristi
    Nov 23 at 7:30















up vote
1
down vote













curr = head;
end = head;
currprev = nullptr;

std::vector<Node*> flag;
while(end) {
sum += end->data;
if (sum < 5) {
flag.push_back(end);
end = end->next;
}

if (sum > 5) {
currprev = curr;
curr = curr->next;
end = curr;
sum = 0;
empty_flag()
}

if (sum == 5) {
curr = end->next;
end = curr;
sum = 0;
delete_nodes_from_flag_from_linked_list()
empty_flag()
}


}



Here is a possible solution. flag is in this case a vector of nodes which are saved until the sum is proven to be 5 or greater as 5. With flag there are two operations which need to be made: empy_flag() - deletes all entries from the vector and delete_nodes_from_flag_from_linked_list() - that deletes the elements from flag from the linked list.



Code for delete_nodes_from_flag_from_linked_list() should be as follows:



auto it = flag.end();
it--;
Node* last = *it;

if (currprev)
currprev->next = last->next;
else
head = last->next

for (auto f : flag)
delete f;


where currprev is the element just before curr. You have to keep track of this element from the beginning, unless your linked list elements keep pointers to previous elements. I am updating the code from above. head is the start of the list.






share|improve this answer























  • how do I implement this? delete_nodes_from_flag_from_linked_list()
    – Benjamin Yakobi
    Nov 22 at 16:27








  • 1




    Please see revised anwer.
    – Cristi
    Nov 22 at 16:53










  • Could you please mark as anwer if it is a correct answer ?
    – Cristi
    Nov 23 at 7:30













up vote
1
down vote










up vote
1
down vote









curr = head;
end = head;
currprev = nullptr;

std::vector<Node*> flag;
while(end) {
sum += end->data;
if (sum < 5) {
flag.push_back(end);
end = end->next;
}

if (sum > 5) {
currprev = curr;
curr = curr->next;
end = curr;
sum = 0;
empty_flag()
}

if (sum == 5) {
curr = end->next;
end = curr;
sum = 0;
delete_nodes_from_flag_from_linked_list()
empty_flag()
}


}



Here is a possible solution. flag is in this case a vector of nodes which are saved until the sum is proven to be 5 or greater as 5. With flag there are two operations which need to be made: empy_flag() - deletes all entries from the vector and delete_nodes_from_flag_from_linked_list() - that deletes the elements from flag from the linked list.



Code for delete_nodes_from_flag_from_linked_list() should be as follows:



auto it = flag.end();
it--;
Node* last = *it;

if (currprev)
currprev->next = last->next;
else
head = last->next

for (auto f : flag)
delete f;


where currprev is the element just before curr. You have to keep track of this element from the beginning, unless your linked list elements keep pointers to previous elements. I am updating the code from above. head is the start of the list.






share|improve this answer














curr = head;
end = head;
currprev = nullptr;

std::vector<Node*> flag;
while(end) {
sum += end->data;
if (sum < 5) {
flag.push_back(end);
end = end->next;
}

if (sum > 5) {
currprev = curr;
curr = curr->next;
end = curr;
sum = 0;
empty_flag()
}

if (sum == 5) {
curr = end->next;
end = curr;
sum = 0;
delete_nodes_from_flag_from_linked_list()
empty_flag()
}


}



Here is a possible solution. flag is in this case a vector of nodes which are saved until the sum is proven to be 5 or greater as 5. With flag there are two operations which need to be made: empy_flag() - deletes all entries from the vector and delete_nodes_from_flag_from_linked_list() - that deletes the elements from flag from the linked list.



Code for delete_nodes_from_flag_from_linked_list() should be as follows:



auto it = flag.end();
it--;
Node* last = *it;

if (currprev)
currprev->next = last->next;
else
head = last->next

for (auto f : flag)
delete f;


where currprev is the element just before curr. You have to keep track of this element from the beginning, unless your linked list elements keep pointers to previous elements. I am updating the code from above. head is the start of the list.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 22 at 18:08

























answered Nov 22 at 15:12









Cristi

1901315




1901315












  • how do I implement this? delete_nodes_from_flag_from_linked_list()
    – Benjamin Yakobi
    Nov 22 at 16:27








  • 1




    Please see revised anwer.
    – Cristi
    Nov 22 at 16:53










  • Could you please mark as anwer if it is a correct answer ?
    – Cristi
    Nov 23 at 7:30


















  • how do I implement this? delete_nodes_from_flag_from_linked_list()
    – Benjamin Yakobi
    Nov 22 at 16:27








  • 1




    Please see revised anwer.
    – Cristi
    Nov 22 at 16:53










  • Could you please mark as anwer if it is a correct answer ?
    – Cristi
    Nov 23 at 7:30
















how do I implement this? delete_nodes_from_flag_from_linked_list()
– Benjamin Yakobi
Nov 22 at 16:27






how do I implement this? delete_nodes_from_flag_from_linked_list()
– Benjamin Yakobi
Nov 22 at 16:27






1




1




Please see revised anwer.
– Cristi
Nov 22 at 16:53




Please see revised anwer.
– Cristi
Nov 22 at 16:53












Could you please mark as anwer if it is a correct answer ?
– Cristi
Nov 23 at 7:30




Could you please mark as anwer if it is a correct answer ?
– Cristi
Nov 23 at 7:30


















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