C++ std::search_n用法及代码示例

先决条件：std::searchstd::search_n是头文件中定义的STL算法，用于搜索给定元素是否满足给定的谓词(如果没有定义这样的谓词则相等)给定的否。与容器元素连续的次数。

它在范围 [first, last) 中搜索计数元素序列，每个元素比较等于给定值(版本 1)或满足谓词(版本 2)。

如果没有找到这样的序列，该函数返回一个迭代器到第一个这样的元素，或一个迭代器到容器的最后一个元素。

std::search_n 的两个版本定义如下 -

1. 使用 == 比较元素：

   用法:

   
   
   
   

    ForwardIterator search_n (ForwardIterator first, ForwardIterator last,
                              Size count, const T& val);
   
   first:
   Forward iterator to beginning of the container to be searched into.
   last:
   Forward iterator to end of the container to be searched into.
   count:
   Minimum number of successive elements to match.
   Size shall be (convertible to) an integral type.
   val:Individual value to be compared.
   返回：
   It returns an iterator to the first element of the sequence.
   If no such sequence is found, the function returns last.
   

   
   // C++ program to demonstrate the use of std::search_n
     
   #include <iostream>
   #include <vector>
   #include <algorithm>
   using namespace std;
   int main()
   {
       int i, j;
     
       // Declaring the sequence to be searched into
       vector<int> v1 = { 1, 2, 3, 4, 5, 3, 3, 6, 7 };
     
       // Declaring the value to be searched for
       int v2 = 3;
     
       // Declaring an iterator for storing the returning pointer
       vector<int>::iterator i1;
     
       // Using std::search_n and storing the result in
       // iterator i1
       i1 = std::search_n(v1.begin(), v1.end(), 2, v2);
     
       // checking if iterator i1 contains end pointer of v1 or not
       if (i1 != v1.end()) {
           cout << "v2 is present consecutively 2 times at index "
                << (i1 - v1.begin());
       } else {
           cout << "v2 is not present consecutively 2 times in "
                << "vector v1";
       }
     
       return 0;
   }

   输出：

   v2 is present consecutively 2 times at index 5
   

2. 使用谓词比较元素：
   用法:

   ForwardIterator search_n ( ForwardIterator first, ForwardIterator last,
                              Size count, const T& val, BinaryPredicate pred );
   
   All the arguments are same as previous template, just one more argument is added
   
   pred: Binary function that accepts two arguments 
   (one element from the sequence as first, and val as second),
    and returns a value convertible to bool. The value returned indicates whether 
   the element is considered a match in the context of this function.
   The function shall not modify any of its arguments. 
   This can either be a function pointer or a function object.
   
   返回：
   It also returns value as per the previous version, i.e., 
   an iterator to the first element of the sequence, 
   satisfying a condition with respect to a given value.
   If no such sequence is found, the function returns last.

   
   // C++ program to demonstrate the use of std::search_n
   // with binary predicate
   #include <iostream>
   #include <vector>
   #include <algorithm>
   using namespace std;
     
   // Defining the BinaryPredicate function
   bool pred(int i, int j)
   {
       if (i == j) {
           return 1;
       } else {
           return 0;
       }
   }
     
   int main()
   {
       int i, j;
     
       // Declaring the sequence to be searched into
       vector<int> v1 = { 1, 2, 3, 4, 5, 3, 3, 6, 7 };
     
       // Declaring the value to be compared to v1 based
       // on a given predicate
       int v2 = 3;
     
       // Declaring an iterator for storing the returning pointer
       vector<int>::iterator i1;
     
       // Using std::search_n and storing the result in
       // iterator i1
       i1 = std::search_n(v1.begin(), v1.end(), 2, v2, pred);
     
       // checking if iterator i1 contains end pointer of v1 or not
       if (i1 != v1.end()) {
           cout << "v2 is present consecutively 2 times at index "
                << (i1 - v1.begin());
       } else {
           cout << "v2 is not present consecutively 2 times "
                << "in vector v1";
       }
     
       return 0;
   }

   输出：

   v2 is present consecutively 2 times at index 5