Array Polygon Questions

Array Polygon Questions

#include <iostream>using namespace std;//array summation int sum (int *a, int n) {return n = = 0? 0:sum (A, n-1) + a[n-1];} Recursive array summation int sum2 (int *a, int n) {int s = 0;for (int i = n-1; i > 0; i--) {s + = A[i];} return s;} Find the maximum minimum value of the array void max_min (int *a, int n, int &max_value, int &min_value) {max_value = A[0];min_value = a[0];for (int i = 0; I < n; i++) {if (A[i] > Max_value) max_value = A[i];if (A[i] < Min_value) Min_value = A[i];}} element int Halftime_val (int *a, int n) with more than half occurrences in the array {typedef struct VALUETYPE{INT num;int time;} Value;int Target_val = -1;value* data = new Value[n];for (int i = 0; i < n; i++)//statistics array elements occurrences {data[i].num = A[i];d ata[ I].time = 1;for (int j = 0; J < i; J + +) {if (Data[i].num = = Data[j].num) data[i].time++;}} for (int i = 0; i < n; i++) {if (Data[i].time > N/2) {target_val = Data[i].num;}} Delete Data;return Target_val;} Find the element in the array with more than half occurrences int find (int* a, int n) {int curvalue = A[0];int count = 1;for (int i = 1; i < n; ++i) {if (a[i] = = CU RValue) COUnt++;else{count--;if (Count < 0) {Curvalue = A[i];count = 1;}}} return curvalue;} Find the smallest two points in the array//given an integer array of n elements, finding out the two elements in the array x and y make the ABS (x-y) value min//First sort, then iterate int compare (const void* A, const void* b) { Return (* (int*) A-* (int*) b);} void min_distance (int *a, int n) {qsort (A, n, sizeof (int), compare); int min_distace = 0;int temp = 0, x1 = 0, x2 = 0;min_di Stace = a[1]-a[0];for (int i = 1; i < n-1; i++) {temp = ABS (A[i + 1]-a[i]); if (temp < min_distace) {Min_distace = Temp;x1 = a[i + 1];x2 = A[i];}} cout << min_distace << endl;cout << x1 << "<< x2 << Endl;} Find the common elements of two ordered arrays */* Given two ordered (non-descending) integer arrays of n elements A and B, find common elements such as a = 0, 1, 2, 3, 4b = 1, 3, 5, 7, 9 output 1, 3***************************** Analysis: Make full use of the order of the array of properties, with two pointers I and J respectively to A and B, compare A[i] and b[j], according to the comparison of the results of moving pointers, there are the following three cases 1. A[i] < B[j], then I increased by 1 and continued to compare 2. A[i] = = B[j], then I and J both add 1, continue to compare 3. A[i] < B[j], then J plus 1, continue to compare the above process until I or J reaches the end of the array. */void find_commom1 (int *a, int *b, int n) {int i = 0, j = 0;while (i < n && J < N) {if (A[i] < b[j]) i++;if (a[i] = = B[j]) {cout << a[i] << ""; i++;j++;} if (A[i] > B[j]) j + +;} cout << Endl;} void Find_commom (int* A, int *b, int n) {int I, j;for (i = 0; i < n; i++) {for (j = 0; J < N; j + +) {if (a[i] = = B[j]) CO UT << a[i] << "";}} cout << Endl;} int main () {int a[4] = {2, 4, 6, 8};int b[5] = {1, 5, 7, 9}; Find_commom (A, B, 4); Find_commom1 (A, B, 4);}

From: http://www.cnblogs.com/graphics/archive/2010/08/24/1761620.html#

Array Polygon Questions