unordered_multimap :: count()是C++ STL中的内置函数,该函数返回容器中其键等于在参数中传递的键的元素数。
句法:
unordered_multimap_name.count(key)
参数:该函数接受单个强制性参数键,该键指定要返回其在unordered_multimap容器中的计数的键。
返回值:返回一个无符号整数类型,该整数类型表示密钥在容器中出现的次数。
下面的程序说明了上述函数:
程序1:
// C++ program to illustrate the
// unordered_multimap::count()
#include
using namespace std;
int main()
{
// declaration
unordered_multimap sample;
// inserts key and element
sample.insert({ 10, 100 });
sample.insert({ 10, 100 });
sample.insert({ 20, 200 });
sample.insert({ 30, 300 });
sample.insert({ 30, 150 });
cout << "10 occurs " << sample.count(10)
<< " times";
cout << "\n20 occurs " << sample.count(20)
<< " times";
cout << "\n13 occurs " << sample.count(13)
<< " times";
cout << "\n30 occurs " << sample.count(30)
<< " times";
return 0;
}
输出:
10 occurs 2 times
20 occurs 1 times
13 occurs 0 times
30 occurs 2 times
程式2:
// C++ program to illustrate the
// unordered_multimap::count()
#include
using namespace std;
int main()
{
// declaration
unordered_multimap sample;
// inserts key and element
sample.insert({ 'a', 'b' });
sample.insert({ 'a', 'b' });
sample.insert({ 'b', 'c' });
sample.insert({ 'r', 'a' });
sample.insert({ 'r', 'b' });
cout << "a occurs " << sample.count('a')
<< " times";
cout << "\nb occurs " << sample.count('b')
<< " times";
cout << "\nz occurs " << sample.count('z')
<< " times";
cout << "\nr occurs " << sample.count('r')
<< " times";
return 0;
}
输出:
a occurs 2 times
b occurs 1 times
z occurs 0 times
r occurs 2 times
要从最佳影片策划和实践问题去学习,检查了C++基础课程为基础,以先进的C++和C++ STL课程基础加上STL。要完成从学习语言到DS Algo等的更多准备工作,请参阅“完整面试准备课程” 。