c++之STL(13) STL 算法 - 查找算法（2）

search_n() 用来查找连续的n个匹配的数值 或者 加谓词

search_n(b, e, c, v)

search_n(b, e, c, v, p)

注意：该方法的第二种形式应该是search_n_if(b, e, c, p)

#include<iostream>
#include<algorithm>
#include<deque>
//
#include<functional>
//
using namespace std;

int main()
{
	deque<int> ideq;
	for (int i = 1; i <= 9; i++)
	{
		if (i == 3)
		{
			ideq.push_back(i);
			ideq.push_back(i);
			ideq.push_back(i);
			ideq.push_back(i);
		}
		else
		{
			ideq.push_back(i);

		}
	}

	for (deque<int>::iterator iter = ideq.begin(); iter != ideq.end(); iter++)
		cout << *iter << endl;
	/*if (pos != string::npos)
		cout << "找到了！" << pos << endl;
	else
		cout << "没找到！" << endl;*/
	deque<int>::iterator pos;
	pos = search_n(ideq.begin(), ideq.end(), 4, 3);
	if (pos != ideq.end())
	{
		cout << "找到了连续的4个3！，序号位置" << distance(ideq.begin(), pos) + 1 << endl;
	}
	else
	{
		cout << "没找到！" << endl;
	}
	// 这里的谓词是 二元 的
	pos = search_n(ideq.begin(), ideq.end(), 3, 6, greater<int>());
	//pos = search_n_if(ideq.begin(), ideq.end(), 3, bind2nd(greater<int>(), 6));

	if (pos != ideq.end())
	{
		cout << "找到了连续的3个大于6的数！start with" << distance(ideq.begin(), pos) + 1 << endl;
	}
	else
		cout << "没找到！" << endl;

	//
	system("pause");
	return 0;
}