Code Implementation and Application of Circular link list
1 Construction
1.1 Background
The Joseph ring problem: Given n persons (Numbers 1,2,3, … , n) sit around a round table and count clockwise from the person named k to the person named m. His next guy starts at 1 again, starts counting clockwise, and the guy who counts to m goes out again; The game continued, until there was only one person left at the round table.
The problem is definitely an application of circular link list, before solving the problem, we first construct the circular link list.
1.2 Construct a circular link list
typedef struct node{
int number;
struct node * next;
}person;
person * initLink(int n){
person * head=(person*)malloc(sizeof(person));
head->number=1;
head->next=NULL;
person * cyclic=head;
for (int i=2; i<=n; i++) {
person * body=(person*)malloc(sizeof(person));
body->number=i;
body->next=NULL;
cyclic->next=body;
cyclic=cyclic->next;
}
cyclic->next=head; //End to end
return head;
}
We can see the only difference from primary link list is that it links end to end, achieving by: cyclic->next=head
2 Solve the problem
void findAndKillK(person * head,int k,int m){
person * tail=head;
while (tail->next!=head) { // Find the node for deletion
tail=tail->next;
}
person * p=head;
while (p->number!=k) { // Find the person with number k
tail=p;
p=p->next;
}
while (p->next!=p) { // Only when p->next==p could we know there's one man left
for (int i=1; i<m; i++) {
tail=p;
p=p->next;
}
tail->next=p->next; // Put down 'p'
printf("Dequeued person's number:%d\n",p->number);
free(p);
p=tail->next; // Continue
}
printf("Dequeued person's number:%d\n",p->number);
free(p);
}
int main() {
printf("The total number n?");
int n;
scanf("%d",&n);
person * head=initLink(n);
printf("Start from the number k?",n);
int k;
scanf("%d",&k);
printf("The dequene number m?");
int m;
scanf("%d",&m);
findAndKillK(head, k, m);
return 0;
}
We assume that there are 6 people in the game, and we start from 4, the person who count to 3 will out and the game continue from next person.
Here’s the output: