這個程式是用位向量來實現的集合，用陣列的下標來表示集合中的元素，而陣列的元素0或1來表示集合有無該元素，因此，它適合於整數集合，且值不是很大的情況。

#include <iostream> #include <assert.h> using namespace std; const int DefaultSize = 100; class Set { private: int * bitVector; int MaxSize; public: Set ( int MaxSetSize = DefaultSize ); ~Set ( ) { delete [ ] bitVector; } void MakeEmpty ( ) { for ( int i = 0; i < MaxSize; i++ ) bitVector[i] = 0; } int AddMember ( const int x ); int DelMember ( const int x ); Set& operator = ( Set & right ); Set& operator + ( Set & right ); Set& operator * ( Set & right ); Set& operator - ( Set & right ); int Contains ( const int x ); int SubSet ( Set & right ); int operator == ( Set & right ); void Print(); }; //建構函式 Set :: Set (int MaxSetSize) : MaxSize (MaxSetSize) { assert ( MaxSize > 0 ); bitVector = new int [MaxSize]; assert ( bitVector != 0 ); for ( int i = 0; i < MaxSize; i++ ) bitVector[i] = 0; } //新增元素，成為返回1,失敗返回0 int Set :: AddMember ( const int x ) { assert ( x >= 0 && x < MaxSize ); if ( ! bitVector[x] ) { bitVector[x] = 1; return 1; } return 0; } //刪除元素，成功返回1,失敗返回0 int Set :: DelMember ( const int x ) { assert ( x >= 0 && x < MaxSize ); if ( bitVector[x] ) { bitVector[x] = 0; return 1; } return 0; } //過載賦值運算子 Set& Set :: operator = ( Set& right ) { assert ( MaxSize == right.MaxSize ); for ( int i =0; i <MaxSize; i++ ) bitVector[i] = right.bitVector[i]; return *this; } //過載加法運算子，表示集合a+b ，並集 Set& Set :: operator + ( Set& right ) { assert ( MaxSize == right.MaxSize ); Set* temp = new Set; for ( int i = 0; i < MaxSize; i++ ) (*temp).bitVector[i] = bitVector[i] || right.bitVector[i]; //有一個為1,即為1 return *temp; } //過載*運算子，表示集合交集 Set& Set :: operator * ( Set & right ) { assert ( MaxSize == right.MaxSize ); Set* temp = new Set; for ( int i = 0; i < MaxSize; i++) (*temp).bitVector[i] = bitVector[i] && right.bitVector[i]; //兩個都為1,才為1 return *temp; } //過載-運算子，表示集合運算a-b Set& Set :: operator - ( Set & right ) { assert ( MaxSize == right.MaxSize ); Set* temp = new Set; for ( int i = 0; i < MaxSize; i++ ) (*temp).bitVector[i] = bitVector[i] && !right.bitVector[i]; //a有，b沒有，才為1 return *temp; } //判斷元素是否在集合中 int Set :: Contains ( const int x ) { assert ( x >= 0 && x < MaxSize ); return bitVector[x]; } // 過載相等判斷 int Set :: operator == ( Set & right ) { assert ( MaxSize == right.MaxSize ); for ( int i = 0; i < MaxSize; i++) if ( bitVector[i] != right.bitVector[i] ) return 0; return 1; } // 首次匹配 int Set :: SubSet (Set & right ) { // assert ( MaxSize == right.MaxSize ); int i; for ( i = 0; i < MaxSize; i++) if (bitVector[i] && right.bitVector[i]) return i; return -1; } void Set::Print() { int i; for(i = 0; i < MaxSize; i++) { if(bitVector[i] == 1) cout << i<<" "; } cout << endl; } int main() { Set s1, s2, s3, s4, s5; int index, equal; //構造集合 for ( int k = 0; k < 10; k++ ) //集合賦值 { s1.AddMember( k ); s2.AddMember( k +7 ); } // s1 : { 0, 1, 2, …, 9 }, s2 : { 7, 8, 9, …, 16 } s3 = s1 + s2; //求s1與s2的並 { 0, 1, …, 16 } // s1.Print(); s4 = s1 * s2; //求s1與s2的交 { 7, 8, 9 } s5 = s1 - s2; //求s1與s2的差 { 0, 1, 2, 3, 4, 5, 6 } index = s2.SubSet ( s5); //求s4在s1中首次匹配 cout << index << endl; //位置， index = 7 equal = s1 == s2; //集合s1與s2比較相等 cout << equal << endl; //equal為0, 兩集合不等 s1.Print(); //cout << endl; s2.Print(); s3.Print(); s4.Print(); s5.Print(); system("pause"); return 0; }