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Doubly linked list concept Node structure Insertion sort Insertion sort program with a doubly linked list

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Head 0abcde0

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7 struct node { node * prev; int val; node * next; };

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1.Ask the use to enter an integer 2.Store the # in one node of a doubly linked list 3.Repeat steps 1 & 2 until the user enters s to terminate input 4.Use the insertion sort to sort the #s that are in the doubly linked list 5.Print out the sorted list

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Head Fig. 1 Fig. 2

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1) Place a pointer called out on the node that we want to insert. (When we 1 st start, out is placed on the next to the last node. 2) Place a pointer called in one node to the right of out. 3) Compare the # to be inserted with the contents of what is pointed to by in. 4) If the # to be sorted is less than what is pointed to by in, swap the contents and move in, one node to the right 5) When in is at the end of the list or the # to be sorted is > than what is pointed to in, we have inserted the # into the correct position and its time to move out, 1 node to the left and repeat steps 2-5.

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1.Well use the insertion sort to rearrange the following list of #s in descending order: Place a pointer called out on the next to the last # 3.Place a pointer called in, one node to the right of out out in 4.Consider the last # (8) to be the sorted list & 3 is the # we want to insert into the sorted list. 5.If the # pointed to by out (3) < the # pointed to by in, swap them. Now you get the following picture out in

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6.After swapping the #s, move in one node to the right out in 7. When in is off the list, this means that we have inserted the # 3 in the correct position 8.Move out, 1 node to the left and place in 1 node to the right of out out in 9.All the nodes to the right of out are sorted in descending order. Now we are going to insert 7 into the list 10.Since 7 is < 8, we need to swap the numbers and also move in one node to the right out in

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11..Compare 7 and 3. 7 is > 3 so we have inserted 7 in to the correct position in the list out in 12. The #s from out and to the right are now sort in descending order. 8, 7,3. Next step is to move out 1 node to the left and place in 1 node to the right of out out in 13. We are going to insert 2 in to the sorted list. You can see that in keeps moving to the right until the # is in the correct position or in is off the list. Out always moves to the left and points to the # we want to insert into the list. This procedure is continues until out points to a null. Then, the list is in descending order.

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#include using std::cin; using std::cout; using std::flush; using std::endl; #include struct node { node* prev; int value; node* next; }; void printList(const node*);

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int main() { char str[15]; node* head = new node; node* tail = head; head >prev = 0; cout<<"enter a number"; cin>>str; while(str[0]!='s') { tail->value=atoi (str); tail >next=new node; tail >next >prev=tail; tail=tail >next; cout<<"enter a number"; cin>>str; } tail >next=0; printList(head);//print unsorted list

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node* in; node* out; int temp; out=tail >prev >prev; while(out!=0) { temp=out >value; in=out >next; while(in >next!=0&&temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } printList(head); // print list return 0; }

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void printList(const node* h) { for (const node* p = h; p; p = p->next) { cout << "node address: " << p << prev << p->prev value next << endl; } }

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#include using std::cin; using std::cout; using std::flush; using std::endl; #include struct node { node* prev; int value; node* next; }; void printList(const node*);

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int main() { char str[15]; node* head = new node; node* tail = head; head >prev = 0; cout<<"enter a number"; cin>>str; head 0 tail str 5

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while(str[0]!='s') { tail->value=atoi (str); tail >next=new node; tail >next >prev=tail; tail=tail >next; cout<<"enter a number"; cin>>str; } tail >next=0; printList(head);//print unsorted list head 05 tail str 5

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while(str[0]!='s') { tail->value=atoi (str); tail >next=new node; tail >next >prev=tail; tail=tail >next; cout<<"enter a number"; cin>>str; } tail >next=0; printList(head);//print unsorted list head 05 tail str 5

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while(str[0]!='s') { tail->value=atoi (str); tail >next=new node; tail >next >prev=tail; tail=tail >next; cout<<"enter a number"; cin>>str; } tail >next=0; printList(head);//print unsorted list head 05 tail str 2

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Head while(str[0]!='s') { tail->value=atoi (str); tail >next=new node; tail >next >prev=tail; tail=tail >next; cout<<"enter a number"; cin>>str; } tail >next=0; printList(head);//print unsorted list tail

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1) Place a pointer called out on the node that we want to insert. (When we 1 st start out is placed on the next to the last node. 2) Place a pointer called in one node to the right of out. 3) Compare the # to be inserted with the contents of what is pointed to by in. 4) If the # to be sorted is less than what is pointed to by in, swap the contents and move in, one node to the right 5) When in is at the end of the list or the # to be sorted is > than what is pointed to in, we have inserted the # into the correct position and its time to move out, 1 node to the left and repeat steps 2-5.

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while(out!=0) {... while(# to be inserted is in the wrong spot) {... in=in >next; //move in one node to the right } out=out >prev; //move out one node to the left }

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Head node* in; node* out; int temp; out=tail >prev >prev; tail out

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next!=0&&temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail out in 3 temp

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next !=0 && temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail out in 3 temp

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next !=0 && temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail out in 3 temp

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next !=0 && temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail outin 7 temp

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next !=0 && temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail outin 7 temp

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Head while (out!=0) { temp=out >value; in=out >next; while(in >next !=0 && temp value) { in >prev >value=in >value; in >value=temp; in=in >next; } out=out >prev; } tail outin 2 temp

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Head tail outin tail out in tail outin Head 2 temp

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Head tail out in tail out in tail out in Head temp 6

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Head tail out= 0 in 5 temp

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Doubly linked list concept Node structure Insertion sort Insertion sort program with a doubly linked list

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