security9802

Здравейте имам задача с условието: Да се състави функция, която извежда на екрана най-често срещания елемент на списък с начален указател START. Имате ли някакви съвети как да направя функцията за най-често срещания елемент? #include<iostream> using namespace std; struct elem { int key; elem* next; elem* prev; }*start = NULL; void add_b(int n); void print_list(); void add_e(int n); void add_bk(int n, int k); void add_ak(int n, int k); void invert_list(); int del_n(int n); int maxlist(); int main() { int ch, n, k; int br = 0; do { cout << "\n1.Add first"; cout << "\n2.Add last"; cout << "\n3.Add before k"; cout << "\n4.Add after k"; cout << "\n5.Delete n"; cout << "\n6.Invert list"; cout << "\n7.Print list"; cout << "\n8.Max List"; cout << "\nYour ch: "; cin >> ch; if (ch < 6) { cout << "\n n="; cin >> n; } if (ch == 3 || ch == 4) { cout << "\nk= "; cin >> k; } switch (ch) { case 1: {add_b(n); print_list(); break; } case 2: {add_e(n); print_list(); break; } case 3: {add_bk(n, k); print_list(); break; } case 4: {add_ak(n, k); print_list(); break; } case 5: {del_n(n); print_list(); break; } case 6: {invert_list(); print_list(); break; } case 7: {print_list(); break; } case 8: {maxlist(); break; } } } while (ch < 8 && ch>0); } void add_b(int n) { elem* p = start; start = new elem; start->key = n; start->next = p; start->prev = NULL; if (p) p->prev = start; } void print_list() { if (start) { cout << "\nList is: "; elem* p = start; while (p) { cout << p->key << "\t"; p = p->next; } } else cout << "\n Empty list!"; } void add_e(int n) { elem* q = new elem; q->next = NULL; q->key = n; if (start == NULL) { start = q; q->prev = NULL; } else { elem* p = start; while (p->next) p = p->next; p->next = q; q->prev = p; } } void add_bk(int n, int k) { elem* p = start; while (p->key != k && p->next) p = p->next; if (p->key == k) { elem* q = new elem; q->next = p->next; q->prev = p; if (p->next) p->next->prev = q; p->next = q; p->key = n; q->key = k; } else cout << "\nNo K"; } void add_ak(int n, int k) { elem* p = start; while (p->key != k && p->next) p = p->next; if (p->key == k) { elem* q = new elem; q->next = p->next; q->prev = p; if (p->next) p->next->prev = q; p->next = q; q->key = n; } else cout << "\nNo K"; } void invert_list() { elem* p = start, * buf = start; while (p) { buf = p->next; p->next = p->prev; if (buf == NULL) start = p; p = buf; } } int del_n(int n) { if (start) { elem* p = start; while (p->key != n && p->next) p = p->next; if (p->key == n) { if (p->next) p->next->prev = p->prev; if (p->prev) p->prev->next = p->next; if (start->key == n) start = start->next; delete p; return 1; } else//if(pi>key==n) { cout << "\n Non"; return 0; } } else { cout << "\n Empty list"; return 0; } }