c ++ STL_6 algorithm (section)

table of Contents

1. for_each()
2. count() and count_if()
3. find() and find_if()
4. search_n()

---

for_each
for_each(InputIterator beg,InputIterator end ,UnaryProc op)

• For the interval [beg,end]of each element call: op(elem).
• Returns opa copy.
• op()You can change elements.
• Linear complexity

Example 1, a simple output

#include<iostream>
#include<algorithm>
#include <vector>
using namespace std;
void print(int elem)
{
	cout << elem << "(普_调用)";
	return;
}
class Print {
public:
	void operator()(int i) {
		cout << i << "(class调用)";
	}
};
int main()
{
	vector<int> num(10, 1);
	for_each(num.begin(),num.end(),print);
	cout << endl;
	for_each(num.begin(), num.end(), Print());
	return 0;
}

Found that the use of function object class when the need to add brackets, using ordinary function of time, does not require parentheses, and heavy-duty related.

Example 2, changing the value of the container

#include<iostream>
#include<algorithm>
#include <vector>
using namespace std;
void Correct(int &elem)
{
	elem += 100;
	return;
}
class Print {
public:
	void operator()(int i) {
		cout << i << "(class调用)";
	}
};
int main()
{
	vector<int> num(10, 1);
	for_each(num.begin(),num.end(), Correct);
	cout << endl;
	for_each(num.begin(), num.end(), Print());
	return 0;
}

Example 3, the return value of the container

#include<iostream>
#include<algorithm>
#include <vector>
#include <iterator>
using namespace std;
class Correct {
private:
	int sum;
public:
	Correct(int _sum):sum(_sum){ }
	void operator()(int num) {
		sum += num;
	}
	 operator int(){
		 return sum;
	}
};
int main()
{
	vector<int> num(10, 1);
	int pos = for_each(num.begin(),num.end(), Correct(0));
	cout << pos <<  endl;
	copy(num.begin(), num.end(), ostream_iterator<int>(cout," "));
	return 0;
}

---

count() and count_if
Without changing the parameter value passed in
count(InputIterator beg,InputIterator end ,const T& value)
the form of the computing [beg end]elements is equal to valuethe number of values.
count_if(InputIterator beg,InputIterator end ,UnaryPredicate op)
The form of unary predicates result is truethe number of elements that need to determine the function.

#include<iostream>
#include<algorithm>
#include <vector>
#include <iterator>
using namespace std;
bool isEven(int elem) {
	return elem % 2 == 0;
}
int main()
{
	vector<int> num{1,2,3,4,5,6,7,8,9,0,10};
	cout << "equal 3 :" << count(num.begin(), num.end(), 3) << endl;
	cout << "even : "<<count_if(num.begin(), num.end(), isEven) << endl;
	cout << "原序列:" << endl;
	copy(num.begin(), num.end(), ostream_iterator<int>(cout," "));
	return 0;
}

---

find() and find_if()
It does not change the value of the parameter passed.
find(InputIterator beg,InputIterator end ,const T& value)
The Returns [beg end]the element is equal to valuethe iterator value of the first element of.
find_if(InputIterator beg,InputIterator end ,UnaryPredicate op)
Analyzing in unary form returns the first iterator, an equivalent value, is not found is returned end().

#include<iostream>
#include<algorithm>
#include <vector>
#include <iterator>
using namespace std;
bool isEven(int elem) {
	return elem % 2 == 0;
}
int main()
{
	vector<int> num{1,2,3,4,5,6,7,8,9,0,10};
	cout << "equal 3 :" << *find(num.begin(), num.end(), 3) << endl;
	cout << "even : "<<*find_if(num.begin(), num.end(), isEven) << endl;
	cout << "原序列:" << endl;
	copy(num.begin(), num.end(), ostream_iterator<int>(cout," "));
	return 0;
}

---

search_n()
Find the value of n consecutive matches before, does not change the passed parameter value.
search_n(InputIterator beg,InputIterator end ,count,const T& value)
Returns the [beg end]elements is equal to valuethe first value of the iterator consecutive equal elements.
search_n(InputIterator beg,InputIterator end ,count,value,BinaryPredicate op)
Analyzing the form of a binary equivalent thereto returns the first iterator consecutive equal values, is not found is returned end().

#include<iostream>
#include<algorithm>
#include <vector>
#include <iterator>
using namespace std;
int main()
{
	vector<int> num{1,2,3,4,5,6,7,8,9,0,0,10};
	cout << "equal 3 :" << *search_n(num.begin(), num.end(),2, 3) << endl;
	cout << "even : "<<*search_n(num.begin(), num.end(),2,6,greater<int>() )<< endl;
	cout << "原序列:" << endl;
	copy(num.begin(), num.end(), ostream_iterator<int>(cout," "));
	return 0;
}