Set Operations(DS)(15C++)

标签（空格分隔）： 程序设计实验 c++

本人学院

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description

Set operations.

You need finish function.

Require knowledge:

1. Know set operations：include

   intersection(交集）

   union（并集）

   the complement of B with respect to A（B对A的补）

   symmetric difference（对称差）

2. copy constructor

3. overload

Require task:

1. Define a copy constructor

2. Overload four operators

Hint:

找到之前的题目Set，并拷贝你们的部分答案到这里，不需要重新写，但不要拷贝标准答案，有抄袭检查。

file provided

main.cpp

////  main.cpp//  C++////  Created by 李天培 on 16/2/25.//  Copyright © 2016年 lee. All rights reserved.//#include <iostream>#include <algorithm>#include "Set.hpp"void display(int* members, int size) {    std::sort(members, members + size);    std::cout << "{";    for (int i = 0; i < size; i++) {        if (i < size - 1) std::cout << members[i] << ", ";        else std::cout << members[i];    }    std::cout << "}" << std::endl;}int main(int argc, const char * argv[]) {    int test[6];    std::cin >> test[0]    >> test[1]    >> test[2]    >> test[3]    >> test[4]    >> test[5];    // Constructor 1    Set s1 = Set();    display(s1.getMembers(), s1.getSize());    std::cout << "is empty set: " << s1.isEmptySet() << std::endl;    // append func    std::cout << "append: " << s1.append(test[0]) << std::endl;    std::cout << "append: " << s1.append(test[4]) << std::endl;    display(s1.getMembers(), s1.getSize());    // repeat append    std::cout << "append: " << s1.append(test[0]) << std::endl;    display(s1.getMembers(), s1.getSize());    std::cout << "is empty set: " << s1.isEmptySet() << std::endl;    // Constructor 2    Set s2 = Set(test, 5);    // remove func    std::cout << "remove: " << s2.remove(test[0]) << std::endl;    display(s2.getMembers(), s2.getSize());    // repeat append    std::cout << "remove: " << s2.remove(test[0]) << std::endl;    display(s2.getMembers(), s2.getSize());    std::cout << test[5] << " is in set: " << s2.isInSet(test[5]) << std::endl;    std::cout << test[2] << " is in set: " << s2.isInSet(test[2]) << std::endl;    display(s1.getMembers(), s1.getSize());    display(s2.getMembers(), s2.getSize());    Set s3;    s3 = s1 & s2;    display(s3.getMembers(), s3.getSize());    s3 = s1 | s2;    display(s3.getMembers(), s3.getSize());    s3 = s1 - s2;    display(s3.getMembers(), s3.getSize());    s3 = s2 - s1;    display(s3.getMembers(), s3.getSize());    s3 = s1 + s2;    display(s3.getMembers(), s3.getSize());    return 0;}

Set.hpp

////  Set.hpp//  C++////  Created by 李天培 on 16/2/25.//  Copyright © 2016年 lee. All rights reserved.//#ifndef Set_hpp#define Set_hpp#include <stdio.h>#define MAX_MEMBERS 100class Set {private:    int members[MAX_MEMBERS];    int size;public:    // Create an empty set.    Set();    // Create an set with some element.    Set(int* m, int s);    // Copy Constructor    Set(Set const &s);    // append a element to set.    // If element in the set, return false.    // Or insert in set and return true.    bool append(int e);    // remove a element by its value from set.    // If element in the set, then remove it and return true.    // Or return false.    bool remove(int e);    // return true if the set is empty, or return false.    bool isEmptySet();    // return true if the element e is in the set, or return false.    bool isInSet(int e);    // & is intersection of two set    Set operator&(const Set &s);    // | is union of two set    Set operator|(const Set &s);    // A - B is the complement of B with respect to A    Set operator-(const Set &s);    // A + B is their symmetric difference. A + B = (A - B) | (B - A)    Set operator+(const Set &s);    // return set.    int* getMembers();    // return size of set.    int getSize();};#endif /* Set_hpp */

understanding

题目要求我们在之前那题set的基础上,增加
拷贝构造函数
重载& | - + 操作符函数

my answer

#include<iostream>#include<cstring>#include"Set.hpp"using namespace std;    Set::Set() {        size = 0;        members[size] = 0;    }    Set::Set(int* m, int s) {        size = 0;        members[size] = 0;        for (int i = 0; i < s; i++) {            append(m[i]);        }    }    Set::Set(Set const &s) {        int i;        memset(this->members, 0, MAX_MEMBERS*sizeof(i));        this->size = s.size;        for (i = 0; i < s.size; i++) {            this->members[i] = s.members[i];        }    }    bool Set::append(int e) {        if (size + 1 >= MAX_MEMBERS || isInSet(e)) return false;            members[size] = e;            size++;            return true;    }    bool Set::remove(int e) {        bool flag = false;        int i;        for (i = 0; i < size; i++) {            if (e == members[i]) {                flag = true;                size--;                for (int j = i; j < size; j++) {                    members[j] = members[j + 1];                }                i--;            }        }        return flag;    }    bool Set::isEmptySet() {        if (size == 0)            return true;        else            return false;    }    int* Set::getMembers() {        return members;    }    int Set::getSize() {        return size;    }    bool Set::isInSet(int e) {        for (int i = 0; i< size; i++) {            if (e == members[i]) {                return true;            }        }        return false;    }    // & is intersection of two set    Set Set::operator&(const Set &s) {        Set temp = s;        for (int i = 0; i < temp.size; i++) {            if (!isInSet(temp.members[i])) {                temp.remove(temp.members[i]);                i--;            }        }        return temp;    }    // | is union of two set    Set Set::operator|(const Set &s) {        Set temp = s;        for (int i = 0; i < size; i++) {            if (!temp.isInSet(members[i])) {                temp.append(members[i]);            }        }        return temp;    }    // A - B is the complement of B with respect to A    Set Set::operator-(const Set &s) {        Set temp1 = s, temp = *this;        for (int i = 0; i < temp.size; i++) {            if (temp1.isInSet(temp.members[i])) {                temp.remove(temp.members[i]);                i--;            }        }        return temp;    }    // A + B is their symmetric difference. A + B = (A - B) | (B - A)    Set Set::operator+(const Set &s) {        Set temp = s;        temp = (*this - temp) | (temp - *this);        return temp;    }

the standard answer

////  Set.cpp//  C++////  Created by 李天培 on 16/2/25.//  Copyright © 2016年 lee. All rights reserved.//#include "Set.hpp"Set::Set() {    size = 0;}Set::Set(int* m, int s) {    size = 0;    for (int i = 0; i < s ; i++) {        if (!isInSet(m[i])) {            members[size++] = m[i];        }        if (size == MAX_MEMBERS) break;    }}Set::Set(Set const &s) {    size = s.size;    for (int i = 0; i < size; i++) {        members[i] = s.members[i];    }}bool Set::append(int element) {    if (size < MAX_MEMBERS && !isInSet(element)) {        members[size] = element;        size++;        return true;    } else {        return false;    }}bool Set::remove(int element) {    for (int i = 0; i < size; i++) {        if (members[i] == element) {            members[i] = members[--size];            return true;        }    }    return false;}bool Set::isEmptySet() {    return (size == 0) ? true : false;}int* Set::getMembers() {    return members;}int Set::getSize() {    return size;}bool Set::isInSet(int element) {    for (int i = 0; i < size; i++) {        if (members[i] == element) {            return true;        }    }    return false;}Set Set::operator&(const Set &s) {    Set temp;    for (int i = 0; i < s.size; i++) {        if (isInSet(s.members[i])) {            temp.members[temp.size++] = s.members[i];        }    }    return temp;}Set Set::operator|(const Set &s) {    Set temp(*this);    for (int i = 0; i < s.size; i++) {        temp.append(s.members[i]);    }    return temp;}Set Set::operator-(const Set &s) {    Set temp(s);    Set temp2(*this);    for (int i = 0; i < temp2.size; i++) {        if (temp.isInSet(temp2.members[i])) {            temp2.remove(temp2.members[i]);        }    }    return temp2;}Set Set::operator+(const Set &s) {    Set temp(s);    return (*this - temp) | (temp - *this);}

rethinking

注意
1.二元运算符不应改变左右操作数,所以应当返回一个新的对象
2.注意i size的变化.