未验证 提交 0174dad7 编写于 作者: C Christian Clauss 提交者: GitHub

rename Operations on Datastructures -> operations_on_datastructures (#655)

上级 ab12ed4a
#include <iostream>
using namespace std;
//node defined
class node
{
public:
int data;
node *link;
node(int d)
{
data = d;
link = NULL;
}
};
//printing the linked list
void print(node *head)
{
node *current = head;
while (current != NULL)
{
cout << current->data << " ";
current = current->link;
}
cout << endl;
}
//creating the linked list with 'n' nodes
node *createlist(int n)
{
node *head = NULL;
node *t = NULL;
for (int i = 0; i < n; i++)
{
node *temp = NULL;
int num;
cin >> num;
temp = new node(num);
if (head == NULL)
{
head = temp;
t = temp;
continue;
}
if (t->link == NULL)
t->link = temp;
t = temp;
}
return head;
}
//performing selection sort on the linked list in an iterative manner
void my_selection_sort_linked_list(node *&head)
{
node *min = head; //throughout the algorithm 'min' is used to denote the node with min value out of all the nodes left for scanning
//while scanning if we find a node 'X' with value lesser than min,
//then we update the pointers in such a way that 'X' becomes the predecessor of 'min'
node *current = min->link; // 'current' refers to the current node we are scanning
node *previous = min; //'previous' refers to the node that is previous to the current node
node *temp = NULL; // 'temp' in this algo is used to point to the last node of the sorted part of the linked list.
//eg. If at any time instance the state of the linked list is suppose 1->2->5->3->8->NULL
//then, we see that "1->2" is the sorted part of the LL, and therefore temp will be pointing to the last node of the sorted part,i.e,'2'
//We keep on arranging the Linked list in such a way that after each iteration the node with 'min' value is placed at its correct position.
//Eg. Let suppose initially we have 5->4->1->3->2->NULL
//After 1st iteration : 1->4->5->3->2->NULL and so on
while (min->link != NULL) //so that all the nodes are scanned or until there exists a node
{
//pick the first node from the unsorted part and assume that it is the minimum and then start scanning from the next node
while (current != NULL) //suppose you choose the min node to be X, then scan starts from the (X+1)th node until its NULL. current = (X+1)th node and min = X
{
if (current->data < min->data) //if the current node is smaller than the presumed node 'min'
{
if (temp == NULL) //temp stays null for the first iteration, therefore it symbolizes that we are scanning for the first time
{
if (previous == min) //if the 'previous' is pointing to the 'min' node
{
//Update the pointers
head = current; //update the head pointer with the current node
min->link = current->link;
current->link = previous;
min = current;
current = previous->link;
}
else //if the 'previous' is not pointing to the 'min' node
{
//Update the pointers
head = current; //update the head pointer with the current node
previous->link = current->link;
current->link = min;
min = current;
current = previous->link;
}
}
else //if 'temp' is not NULL, i.e., its not the 1st iteration
{
temp->link = current;
previous->link = current->link;
current->link = min;
min = current;
current = previous->link;
}
}
else //if the current node is greater than min, just move the previous and the current pointer a step further
{
previous = previous->link;
current = current->link;
}
}
//update the pointers. Set 'temp' to the last node in the sorted part. Make 'min' move a step further so that 'min' points to the 1st node of the unsorted part
//start the iteration again
temp = min;
min = min->link;
previous = min;
current = min->link;
}
}
// Test cases:
// enter the no. of nodes : 5
// 8 9 3 1 4
// original list is : 8 9 3 1 4
// sorted list is : 1 3 4 8 9
// enter the no. of nodes : 3
// -1 -2 -3
// original list is : -1 -2 -3
// sorted list is : -3 -2 -1
// enter the no. of nodes : 8
// 8 7 6 5 4 3 2 1
// original list is : 8 7 6 5 4 3 2 1
// sorted list is : 1 2 3 4 5 6 7 8
// enter the no. of nodes : 6
// 5 3 4 1 -2 -4
// original list is : 5 3 4 1 -2 -4
// sorted list is : -4 -2 1 3 4 5
int main()
{
node *head = NULL;
int n;
cout << "enter the no. of nodes : "; //taking input from user about the number of nodes in linked list
cin >> n;
if (n == 0)
return 0;
head = createlist(n); //creating the list
cout << "original list is : ";
print(head); //printing the original linked list
my_selection_sort_linked_list(head); //applying selection sort
cout << "sorted list is : ";
print(head); //printing the sorted linked list
return 0;
#include <iostream>
using namespace std;
//node defined
class node
{
public:
int data;
node *link;
node(int d)
{
data = d;
link = NULL;
}
};
//printing the linked list
void print(node *head)
{
node *current = head;
while (current != NULL)
{
cout << current->data << " ";
current = current->link;
}
cout << endl;
}
//creating the linked list with 'n' nodes
node *createlist(int n)
{
node *head = NULL;
node *t = NULL;
for (int i = 0; i < n; i++)
{
node *temp = NULL;
int num;
cin >> num;
temp = new node(num);
if (head == NULL)
{
head = temp;
t = temp;
continue;
}
if (t->link == NULL)
t->link = temp;
t = temp;
}
return head;
}
//performing selection sort on the linked list in an iterative manner
void my_selection_sort_linked_list(node *&head)
{
node *min = head; //throughout the algorithm 'min' is used to denote the node with min value out of all the nodes left for scanning
//while scanning if we find a node 'X' with value lesser than min,
//then we update the pointers in such a way that 'X' becomes the predecessor of 'min'
node *current = min->link; // 'current' refers to the current node we are scanning
node *previous = min; //'previous' refers to the node that is previous to the current node
node *temp = NULL; // 'temp' in this algo is used to point to the last node of the sorted part of the linked list.
//eg. If at any time instance the state of the linked list is suppose 1->2->5->3->8->NULL
//then, we see that "1->2" is the sorted part of the LL, and therefore temp will be pointing to the last node of the sorted part,i.e,'2'
//We keep on arranging the Linked list in such a way that after each iteration the node with 'min' value is placed at its correct position.
//Eg. Let suppose initially we have 5->4->1->3->2->NULL
//After 1st iteration : 1->4->5->3->2->NULL and so on
while (min->link != NULL) //so that all the nodes are scanned or until there exists a node
{
//pick the first node from the unsorted part and assume that it is the minimum and then start scanning from the next node
while (current != NULL) //suppose you choose the min node to be X, then scan starts from the (X+1)th node until its NULL. current = (X+1)th node and min = X
{
if (current->data < min->data) //if the current node is smaller than the presumed node 'min'
{
if (temp == NULL) //temp stays null for the first iteration, therefore it symbolizes that we are scanning for the first time
{
if (previous == min) //if the 'previous' is pointing to the 'min' node
{
//Update the pointers
head = current; //update the head pointer with the current node
min->link = current->link;
current->link = previous;
min = current;
current = previous->link;
}
else //if the 'previous' is not pointing to the 'min' node
{
//Update the pointers
head = current; //update the head pointer with the current node
previous->link = current->link;
current->link = min;
min = current;
current = previous->link;
}
}
else //if 'temp' is not NULL, i.e., its not the 1st iteration
{
temp->link = current;
previous->link = current->link;
current->link = min;
min = current;
current = previous->link;
}
}
else //if the current node is greater than min, just move the previous and the current pointer a step further
{
previous = previous->link;
current = current->link;
}
}
//update the pointers. Set 'temp' to the last node in the sorted part. Make 'min' move a step further so that 'min' points to the 1st node of the unsorted part
//start the iteration again
temp = min;
min = min->link;
previous = min;
current = min->link;
}
}
// Test cases:
// enter the no. of nodes : 5
// 8 9 3 1 4
// original list is : 8 9 3 1 4
// sorted list is : 1 3 4 8 9
// enter the no. of nodes : 3
// -1 -2 -3
// original list is : -1 -2 -3
// sorted list is : -3 -2 -1
// enter the no. of nodes : 8
// 8 7 6 5 4 3 2 1
// original list is : 8 7 6 5 4 3 2 1
// sorted list is : 1 2 3 4 5 6 7 8
// enter the no. of nodes : 6
// 5 3 4 1 -2 -4
// original list is : 5 3 4 1 -2 -4
// sorted list is : -4 -2 1 3 4 5
int main()
{
node *head = NULL;
int n;
cout << "enter the no. of nodes : "; //taking input from user about the number of nodes in linked list
cin >> n;
if (n == 0)
return 0;
head = createlist(n); //creating the list
cout << "original list is : ";
print(head); //printing the original linked list
my_selection_sort_linked_list(head); //applying selection sort
cout << "sorted list is : ";
print(head); //printing the sorted linked list
return 0;
}
\ No newline at end of file
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