search
为序列中的第一个匹配项的搜索。事实上谓词二进制等效的指定这些元素与相等在元素特定顺序或元素与元素位于给定序列的目标范围中。
template<class ForwardIterator1, class ForwardIterator2>
ForwardIterator1 search(
ForwardIterator1 _First1,
ForwardIterator1 _Last1,
ForwardIterator2 _First2,
ForwardIterator2 _Last2
);
template<class ForwardIterator1, class ForwardIterator2, class Predicate>
ForwardIterator1 search(
ForwardIterator1 _First1,
ForwardIterator1 _Last1,
ForwardIterator2 _First2,
ForwardIterator2 _Last2
Predicate _Comp
);
参数
_First1
寻址要搜索的范围中的第一个元素的位置的向前迭代器。_Last1
寻址要搜索的范围中最后一个元素的下一位置的向前迭代器。_First2
处理第一元素的位置将向前迭代器将匹配的范围。_Last2
寻址最终元素的前向迭代器位置一将匹配的范围。_Comp
定义将满足的条件的用户定义的谓词函数对象,如果两个元素将采用为相等。 二进制谓词采用两个参数,并且在满足时返回 true,在未满足时返回 false。
返回值
处理第一 subsequence 的第一个元素的位置实际上谓词二进制等效指定的匹配指定的顺序或向前迭代器。
备注
operator== 用于确定在元素和指定值之间的匹配必须对在其操作数之间存在着用关系。
引用的范围必须是有效的;所有指针必须 dereferenceable,在每个序列中最后位置从开始来访问通过递增。
平均复杂是线性的相对于搜索范围的大小,而最坏情况的复杂,也是与搜索线性序列的范围。
示例
// alg_search.cpp
// compile with: /EHsc
#include <vector>
#include <list>
#include <algorithm>
#include <iostream>
// Return whether second element is twice the first
bool twice (int elem1, int elem2 )
{
return 2 * elem1 == elem2;
}
int main( ) {
using namespace std;
vector <int> v1, v2;
list <int> L1;
vector <int>::iterator Iter1, Iter2;
list <int>::iterator L1_Iter, L1_inIter;
int i;
for ( i = 0 ; i <= 5 ; i++ )
{
v1.push_back( 5 * i );
}
for ( i = 0 ; i <= 5 ; i++ )
{
v1.push_back( 5 * i );
}
int ii;
for ( ii = 4 ; ii <= 5 ; ii++ )
{
L1.push_back( 5 * ii );
}
int iii;
for ( iii = 2 ; iii <= 4 ; iii++ )
{
v2.push_back( 10 * iii );
}
cout << "Vector v1 = ( " ;
for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
cout << *Iter1 << " ";
cout << ")" << endl;
cout << "List L1 = ( " ;
for ( L1_Iter = L1.begin( ) ; L1_Iter!= L1.end( ) ; L1_Iter++ )
cout << *L1_Iter << " ";
cout << ")" << endl;
cout << "Vector v2 = ( " ;
for ( Iter2 = v2.begin( ) ; Iter2 != v2.end( ) ; Iter2++ )
cout << *Iter2 << " ";
cout << ")" << endl;
// Searching v1 for first match to L1 under identity
vector <int>::iterator result1;
result1 = search (v1.begin( ), v1.end( ), L1.begin( ), L1.end( ) );
if ( result1 == v1.end( ) )
cout << "There is no match of L1 in v1."
<< endl;
else
cout << "There is at least one match of L1 in v1"
<< "\n and the first one begins at "
<< "position "<< result1 - v1.begin( ) << "." << endl;
// Searching v1 for a match to L1 under the binary predicate twice
vector <int>::iterator result2;
result2 = search (v1.begin( ), v1.end( ), v2.begin( ), v2.end( ), twice );
if ( result2 == v1.end( ) )
cout << "There is no match of L1 in v1."
<< endl;
else
cout << "There is a sequence of elements in v1 that "
<< "are equivalent\n to those in v2 under the binary "
<< "predicate twice\n and the first one begins at position "
<< result2 - v1.begin( ) << "." << endl;
}
要求
标头: <算法>
命名空间: std