检查两个给定的集合是否不相交？
当两个集合没有公共元素时，它们是不相交的集合。换句话说，如果我们得到两个集合的交集，那么我们将得到空集合。
该方法很简单，该算法给出了两组。我们假设两个集合都已排序，并且在两个集合之间比较了项目。当存在匹配项时，它不是不相交集，当没有项目匹配时，它们就是不相交集。
输入输出
Input:
Two sets:
set1: {15, 12, 36, 21, 14}
set2: {7, 89, 56, 32}
Output:
Both sets are disjoint算法
isDisjoint(set1, set2)
输入：两套。
输出：当两个集合不相交时为真。
Begin
i1 := start of first set
i2 := start of second set
while i1 in set1 and i2 in set 2, do
if set1[i1] < set2[i2], then
i1 := i1 + 1
else if set2[i2] < set1[i1], then
i2 := i2 + 1
else
return false
done
return true
End示例
#include<iostream>
#include<set>
using namespace std;
bool isDisjoint(set<int> set1, set<int> set2) {
set<int>::iterator i1, i2;
i1 = set1.begin(); i2 = set2.begin(); //initialize iterators with first element
while(i1 != set1.end() && i2 != set2.end()) { //when both set have some elements to check
if(*i1 < *i2)
i1++; //when item of first set is less than second set
else if(*i2 < *i1)
i2++; //when item of second set is less than first set
else
return false; //if items are matched, sets are not disjoint
}
return true;
}
int main() {
set<int> set1, set2;
int n1, n2;
cout << "Enter number of elements in set 1: "; cin >>n1;
while(n1 != set1.size()) { //duplicate items will be discarded
int item;
cout << "Enter element: "; cin >> item;
set1.insert(item);
}
cout << "Enter number of elements in set 2: "; cin >>n2;
while(n2 != set2.size()) {
int item;
cout << "Enter element: "; cin >> item;
set2.insert(item);
}
if(isDisjoint(set1, set2))
cout << "Both sets are disjoint";
else
cout << "Sets are not disjoint";
}输出结果
Enter number of elements in set 1: 5 Enter element: 15 Enter element: 12 Enter element: 36 Enter element: 21 Enter element: 14 Enter number of elements in set 2: 4 Enter element: 7 Enter element: 89 Enter element: 56 Enter element: 32 Both sets are disjoint