[Algorithm world] about single-linked list operation has ring-free judgment

#include <stdio.h> #include <stdlib.h>//a variety of function tests for the ring list typedef struct NODE{INT data;struct Node *next;} node;typedef struct node* linklist;/* list initialization */int initlist (linklist *l) {*l = (linklist) malloc (sizeof (Node)); *l)) return-1; (*l)->next = Null;return 0;} /* The length of the list */int listlength (linklist L) {int i = 0; Linklist p = l->next;while (p) {i++;p = P->next;} return i;} int Createlisthead (linklist *l, int n) {linklist p;int i;for (i=0; i<n; i++) {p = (linklist) malloc (sizeof (Node));p data = I+1;p->next = (*l)->next; (*l)->next = p;}} int Createlisttail (linklist *l, int n) {linklist p, r;int i;r = *l;for (i=0; i<n; i++) {p = (linklist) malloc (sizeof (Node)) ;p->data = I+1;r->next = P;r = P;} R->next = NULL;} /* Print List */void listtraverse (linklist L) {linklist p = l->next;while (p) {printf ("%d", p->data);p = P->next;} printf ("\ n");} /* Single linked list reversal */linklist Listreverse (linklist l) {linklist Current, Pnext, prev;if (L = = NULL | | L->next = = NULL) return l;current = L->next;pNext = Current->next;current->next = Null;while (pnext) {prev = Pnext->next;pnext->next = Current;current = PN Ext;pnext = prev;} L->next = Current;return L;} Linklist ListReverse2 (linklist l) {linklist Current, p;if (L = = NULL) return null;current = l->next;printf ("current =%d\       n ", Current->data); while (current->next! = NULL) {/** for example * p = current->next; 9 number words need eight times, the first P equals 8* currnet->next = p->next;       The next value of P is 7, and the next * P->next = l->next; of the current is paid             Assign the first node of L 9 to the next value of P * l->next = p;          The next value of L is 8 because the next of the previous p is already 9*/p = current->next; Current->next = P->next;p->next = l->next; L->next = P;} return L;} int Listinsert (linklist *l, int pos, int num) {int J; Linklist p, s;p = *l;j = 1;while (P && j<pos) {p = p->next;++j;} if (!p | | j>pos) return-1;s = (linklist) malloc (sizeof (Node)); s->data = Num;s->next = P->next;p->next = S;r Eturn 0;} int IsEmpty (linklist L) {if (L->next) return-1;elsereturn 1;} int Buildlistloop (linklist *l, int num) {int i = 0; Linklist cur = (*l)->next; linklist tail= null;if (num <= 0 | | L = = null) return-1;for (i=1; i<num; ++i) {if (cur = = null) {return-1;} cur = cur->next;} Tail = Cur;while (tail->next) {tail = Tail->next;} Tail->next = Cur;return 0;} int Listtraveselimit (linklist L, int n) {int i = 0; Linklist p = l->next;while (P && i<n) {printf ("%d", p->data);p = p->next;i++;} printf ("\ n only shows%d \ n", "Nth"); return 0;} int Hasloop (linklist l) {linklist fast = l; linklist slow = l;/** * here if there is no ring, then fast arrives at the end point * When the list length is odd, fast->next is empty * When the list length is even, fast is empty */while (fast! = NULL && F Ast->next! = NULL) {fast = Fast->next->next;slow = slow->next;if (fast = = slow) break;} if (fast = = NULL | | fast->next = NULL) Return-1;elsereturn 1;} int Looplength (linklist L) {if ( -1 = = Hasloop (l)) return 0; Linklist fast = L; linklist slow = l;int length = 0;bool begin = False;bool again = false;while (fast! = NULL && Fast->neXT = NULL) {fast = Fast->next->next;slow = slow->next;//stops counting after two laps, jumps out of the loop if (fast = = Slow && again = = True ) {break;} After a lap start counting if (fast = = Slow && again = = false) {begin = True;again = true;} Count if (begin = = True) ++length;} return length;}  Solve the problem of entry point ideas//8 O-o 7/////9 o O 6//\///O-o-O-o- o//1 2 3 4 5//The distance from the meeting point to the connection point is equal to the distance from the head pointer to the connection point//First let the speed pointer meet, and then the slow pointer re-points to the head node//from the meeting point, the head pointer to start walking, meet the shop is the connection point linklist Findloopentera CE (linklist l) {linklist fast = L; Linklist slow = l;while (fast! = NULL && Fast->next! = null) {fast = Fast->next->next;slow = Slow->nex t;//If there is a ring, then fast will exceed the slow lap if (fast = = slow) break;} if (fast = = NULL | | fast->next = = NULL) return Null;slow = L;while (slow! = fast) {slow = Slow->next;fast = Fast->nex t;} return slow;} int Getmidnode (linklist l, int *e) {linklist fast = l; Linklist slow = L;while (fast->next! = null) {if (Fast->next->next! = null) {fast = FAST-&GT;next->next;slow = Slow->next;} Else{fast = Fast->next;}} *e = Slow->data;return 0;} int clearlist (linklist *l) {linklist p, q;p = (*l)->next;while (p) {q = P->next;free (p);p = q;} (*l)->next = Null;return 0;} int main () {linklist l;int i;int n;int e;i = initlist (&l); if (i! =-1) printf ("Initlist () initialization succeeded. \ n");p rintf ("After the list L is initialized, Li Stlength (L) =%d\n ", Listlength (L)); for (i=0; i<10; i++) {Listinsert (&l, 1, i);} printf ("IsEmpty =%d (1: null,-1: Non-empty) \ n", IsEmpty (L));p rintf ("Listlength (L) =%d\n", Listlength (L));p rintf ("Print List:"); Listtraverse (L);//reverse reversal of LISTREVERSE2 (l);p rintf ("Reverse Printing chain list:"); Listtraverse (L);//The middle node of the list Getmidnode (L, &e);p rintf ("Intermediate node of the list e=%d\n", E);p rintf ("Ring the list, enter position:"); scanf ("%d", &AMP;N); Buildlistloop (&l, N); Listtraveselimit (L, 20);//Determine if the list has a ring if (1 = = Hasloop (L)) printf ("Chain list has a ring \ n"), elseprintf ("Chain list without ring \ n"),//Calculate ring length printf ("Calculate the ring length, Looplength (L) =%d\n ", Looplength (L));//Find the entry point of the ring linklist enterance = Findloopenterace (l);p rintf (" The value of the entry point is:%d\n ", Enterance->data);//printf ("AdoptedHead interpolation method: \ n ", Createlisthead (&l));//listtraverse (L);//printf (" Using the tail interpolation method \ n ", Createlisttail (&l, 10));// Listtraverse (L);//i = Clearlist (&l);//if (i = = 0)//printf ("Empty linklist list. \ n");//printf ("IsEmpty =%d (1: null,-1: Non-empty) \ n" , IsEmpty (L));//printf ("Listlength (L) =%d\n", Listlength (L)); return 0;}

　　

[Algorithm world] about single-linked list operation has ring-free judgment