Linked list, constructors, macros, pointers

First, how to judge a single linked list is a ring?   (Note that you cannot use the flag bit, up to two extra pointers) to struct node {char val; node* next;}    BOOL Check (const node* head) {}//return false: No ring; True: There is an O (n) method that is (make two pointers, one step at a time, one increment of each step, and if there is a ring, the two must overlap, and vice versa):            BOOL Check (const node* head) {if (head==null) return false;         Node *low=head, *fast=head->next;               while (Fast!=null && fast->next!=null) {low=low->next;               fast=fast->next->next;        if (low==fast) return true;   } return false;    } Second, delete the most intermediate element of a single list, requiring time as short as possible (cannot use two cycles) struct link{int data; struct link *next;};    void Delmiddle (link *head) {if (head = = NULL) return;            else if (Head->next = = NULL) {delete head;    Return            } else {link *low = head;            Link *fast = head->next; while (fast! = NULL && Fast->next! = null) {fast =fast->next->next;                       if (fast = = NULL) break;            Low = low->next;            } Link *temp = low->next;            Low->next = low->next->next;      Delete temp;       }}int Main () {struct link *head,*l;       struct link *s;       Head = (link*) malloc (sizeof);       head->data=0;       Head->next = NULL;       L = head;            for (int i=1; i<9; i++) {s = (link*) malloc (sizeof);            S->data = i;            S->next = NULL;            l->next= s;       L = l->next;       } print (head);       Delmiddle (head);       Print (head); return 0;} Three, enter N, for a n*n matrix, specify the matrix along the 45-degree line increment (via)/** * Get a two-dimensional array of the following style * zigzag (the sequence of pixel data in JPEG encoding) * * 0, 1, 5, 6,14,15,27,28,* 2, 4, 7,13,16,26 , 29,42,* 3, 8,12,17,25,30,41,43,* 9,11,18,24,31,40,44,53,* 10,19,23,32,39,45,52,54,* 20,22,33,38,46,51,55,60,* 2 1,34,37,47,50,56,59,61,* 35,36,48,49,57,58,62,63 */void Zigzag (int n) {int **a = (int**) malloc (n*sizeof (int *)); Allocate space if (NULL = = a) return; int i;                for (i = 0; i < n; i++) {if ((a[i] = (int*) malloc (n * sizeof (int.)) = = NULL) {while (--i>=0)            Free (a[i]);            Free (a);        Return }} bool flag = FALSE; This flag is used to determine whether to generate an int count = 0 from a 45-degree angle or a 225-degree angle;    for (i=0; i<n; i++)//The upper half of the generated data {if (flag) {for (int r = 0; r<=i; r++) {A[r][i-r] = count;   count++;  } flag = false;    } else {for (int r = i; r>=0; r--) {A[r][i-r] = count;   count++;  } flag = true;  }} for (i=n-1; i>=0; i--)//Generate data for the lower half {//cout<<i<<endl;       if (flag) {for (int r = 0; r<=i-1; r++) {int r1 = N-i+r;  Represents the current line int c1 = 2*N-I-1-R1;    Represents the current column a[r1][c1] = count;   count++;  } flag = false;    } else {for (int r = i-1; r>=0; r--) {cout<< "ddd" <<endl;    int r1 = N-i+r; int C1 = 2*N-I-1-R1;    cout<<r1<< "," <<c1<<endl;    A[R1][C1] = count;   count++;  } flag = true;  }} for (int r = 0; r<n; r++) {for (int c=0; c<n; C + +) cout<<a[r][c]<< ","; cout<<endl; }}int Main () {int n; cin>>n; zigzag (n); return 0;}  Another person on the web wrote a more ingenious algorithm:/*** get a two-dimensional array of the following styles * zigzag (the sequence of pixel data in JPEG encoding) * * 0, 1, 5, 6,14,15,27,28,* 2, 4, 7,13,16,26,29,42,* 3, 8,12,17,25,30,41,43,* 9,11,18,24,31,40,44,53,* 10,19,23,32,39,45,52,54,* 20,22,33,38,46,51,55,60,* 21,34,37,47,50    , 56,59,61,* 35,36,48,49,57,58,62,63*/#include <stdio.h>int main () {int N;    int S, I, J;    int Squa;    scanf ("%d", &n);    /* Allocate space */int **a = malloc (N * sizeof (int *));    if (a = = NULL) return 0;                for (i = 0; i < N; i++) {if (a[i] = malloc (N * sizeof (int.)) = = NULL) {while (--i>=0)            Free (a[i]);            Free (a);        return 0;        }}/* Array assignment */Squa = n*n; for (i = 0; i < N;            i++) for (j = 0; J < N; J + +) {s = i + j;            if (S < N) A[i][j] = s* (s+1)/2 + (((i+j)%2 = = 0)? i:j);                else {s = (n-1-i) + (N-1-J);            A[I][J] = squa-s* (s+1)/2-(N-(((i+j)%2 = = 0)? i:j)); }/* Print output */for (i = 0; i < n; i++) {for (j = 0; J < N; j + +) printf ("%-6d", a[i        ][J]);    printf ("\ n"); } return 0;}    Four, printing integers 1 to 1000, cannot use Process Control statements (For,while,goto, etc.) also cannot use recursive 1.typedef struct _test{static int A;        _test () {printf ("%d\n", _test::a);    a++;  }}test;  int test::a = 1;    int main () {Test tt[1000];  return 0; } 2. #include <stdio.h> #define B p,p,p,p,p,p,p,p,p,p #define P l,l,l,l,l,l,l,l,l,l #define L i,i,i,i     , I,i,i,i,i,i,n #define I printf ("%3d", i++) #define N printf ("\ n") int main () {int I = 1; B  } or # define A (x) x;x;x;x;x;x;x;x;x;x;int main () {int n = 1;   A (A (A (printf ("%d", n++))); return 0;}         V. struct S {int i; int * p; };         void Main () {s S;         int * p = &s.i;         P[0] = 4;         P[1] = 3;         S.P = p;         S.P[1] = 1; S.p[0] = 2; Ask the program which line will die.          (Microsoft) Solution: s S;        int * p = &s.i;                    The address of the S.I is stored in p p[0] = 4;                    Modified S.I p[1] = 3;                    Modified the S.P S.P = p;               S.P point s.i s.p[1] = 1;               Modify S.P itself s.p[0] = 2;   S.P point is 0x00000001, try to write here, error s.p[0] = 2; Error because S.P is S.I address, s.p[1] is S.P, when S.p[1]=1, S.P at this time storage is 1, instead of address s.i, so in s.p[0] = 2 o'clock error. At this time the equivalent of s.p=ox00000001; address ox0000001 = 2; Of course it went wrong. If the statement s.p[0] =2 precedes s.p[1]=1, the program will not go wrong. At this time the statement is equivalent to S.i=2;s.p=1;   Six, the title description: 1.     int swap (int *x,int *y) {if (x==null¦¦y==null) return-1;     *x + = *y; *y = *x-*Y       *x-= *y; return 1; Please make a mistake, overflow has been considered, not error 2.     void foo (int *x, int *y) {*x + = *y; *x + = *y;   } void Fun (int *x, int *y) {*x + = 2 * (*y); } ask whether the two functions are equivalent and can be exchanged for answers: The function of the first question is exchange. But if x is considered, y is pointing to the same variable, and the result is that the value of this variable is 0. The two functions of the second question are different, also consider that x, Y is pointing to the same variable. The result of the first function is 4 times times that of the variable. But the result of the second function is 3 times times the variable.

Linked list, constructors, macros, pointers