C++ STL 算法精选之查找篇

1.查找类算法
adjacent_find(first,last);

查找区间[first,last)内第一次出现连续的两个相等的元素，并返回指向第一个元素的迭代器，连续元素之间的比较，默认是==

adjacent_find(first,last,pred);

用途如上，但是元素之间的比较是通过函数pred来完成，pred接受两个容器内元素类型的元素，返回bool值

函数原型：
template <class ForwardIterator>
   ForwardIterator adjacent_find ( ForwardIterator first, ForwardIterator last )
{
  ForwardIterator next=first; ++next;
  if (first != last)
    while (next != last)
      if (*first++ == *next++)  // or: if (pred(*first++,*next++)), for the pred version
        return first;
  return last;
}

 例子：
// adjacent_find example
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

bool myfunction (int i, int j) {
  return (i==j);
}

int main () {
  int myints[] = {10,20,30,30,20,10,10,20};
  vector<int> myvector (myints,myints+8);
  vector<int>::iterator it;

  // using default comparison:
  it = adjacent_find (myvector.begin(), myvector.end());

  if (it!=myvector.end())
    cout << "the first consecutive repeated elements are: " << *it << endl;

  //using predicate comparison:
  it = adjacent_find (++it, myvector.end(), myfunction);

  if (it!=myvector.end())
    cout << "the second consecutive repeated elements are: " << *it << endl;
 
  return 0;
}
 

Output:
the first consecutive repeated elements are: 30
the second consecutive repeated elements are: 10

 find(first,last,value);
 查找区间[first,last)之间内值为value的元素，返回迭代器类型，若没找到，则返回迭代器末尾end 函数原型：
 template<class InputIterator, class T>
  InputIterator find ( InputIterator first, InputIterator last, const T& value )
  {
    for ( ;first!=last; first++) if ( *first==value ) break;
    return first;
  }
例子：// find example
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main () {
  int myints[] = { 10, 20, 30 ,40 };
  int * p;

  // pointer to array element:
  p = find(myints,myints+4,30);
  ++p;
  cout << "The element following 30 is " << *p << endl;

  vector<int> myvector (myints,myints+4);
  vector<int>::iterator it;

  // iterator to vector element:
  it = find (myvector.begin(), myvector.end(), 30);
  ++it;
  cout << "The element following 30 is " << *it << endl;

  return 0;
}
 

Output:
The element following 30 is 40
The element following 30 is 40

find_if(first,last,pred);
返回区间[first,last)内第一个使pred函数返回为真的元素的迭代器，否则返回last注意：pred接受一个参数函数原型：
template<class InputIterator, class Predicate>
  InputIterator find_if ( InputIterator first, InputIterator last, Predicate pred )
  {
    for ( ; first!=last ; first++ ) if ( pred(*first) ) break;
    return first;
  }
例子：// find_if example
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

bool IsOdd (int i) {
  return ((i%2)==1);
}

int main () {
  vector<int> myvector;
  vector<int>::iterator it;

  myvector.push_back(10);
  myvector.push_back(25);
  myvector.push_back(40);
  myvector.push_back(55);

  it = find_if (myvector.begin(), myvector.end(), IsOdd);
  cout << "The first odd value is " << *it << endl;

  return 0;
}
 

Output:
The first odd value is 25

find_first_of(first1,last1,first2,last2);
find_first_of(first1,last1,first2,last2,pred);
返回一个迭代器，使得[first2,last2)之中任意一个元素第一次出现在区间[first1,last1)中。
默认比较为==,当然也可以自己定义pred函数(接受2个参数)，返回bool型 函数原型：
template<class ForwardIterator1, class ForwardIterator2>
  ForwardIterator1 find_first_of ( ForwardIterator1 first1, ForwardIterator1 last1,
                                   ForwardIterator2 first2, ForwardIterator2 last2)
{
  for ( ; first1 != last1; ++first1 )
    for (ForwardIterator2 it=first2; it!=last2; ++it)
      if (*it==*first1)          // or: if (comp(*it,*first)) for the pred version
        return first1;
  return last1;
}
例子：// find_first_of example
#include <iostream>
#include <algorithm>
#include <cctype>
#include <vector>
using namespace std;

bool comp_case_insensitive (char c1, char c2) {
  return (tolower(c1)==tolower(c2));
}

int main () {
  int mychars[] = {'a','b','c','A','B','C'};
  vector<char> myvector (mychars,mychars+6);
  vector<char>::iterator it;

  int match[] = {'A','B','C'};

  // using default comparison:
  it = find_first_of (myvector.begin(), myvector.end(), match, match+3);

  if (it!=myvector.end())
    cout << "first match is: " << *it << endl;

  // using predicate comparison:
  it = find_first_of (myvector.begin(), myvector.end(),
                      match, match+3, comp_case_insensitive);

  if (it!=myvector.end())
    cout << "first match is: " << *it << endl;
 
  return 0;
}
 

Output:
First match is: A
First match is: a

find_end(first1,last1,first2,last2);
find_end(first1,last1,first2,last2,pred);
返回一个元素迭代器，使得在区间[first1,last1)中最后一次出现[fiest2,last2),
默认比较为==，当然也可以写自己的比较函数pred,接受两个参数，返回bool值函数原型：
template<class ForwardIterator1, class ForwardIterator2>
  ForwardIterator1 find_end ( ForwardIterator1 first1, ForwardIterator1 last1,
                              ForwardIterator2 first2, ForwardIterator2 last2)
{
  ForwardIterator1 it1, limit, ret;
  ForwardIterator2 it2;

  limit=first1; advance(limit,1+distance(first1,last1)-distance(first2,last2));
  ret=last1;

  while (first1!=limit)
  {
    it1 = first1; it2 = first2;
    while (*it1==*it2)          // or: while (pred(*it1,*it2)) for the pred version
      { ++it1; ++it2; if (it2==last2) {ret=first1;break;} }
    ++first1;
  }
  return ret;
}
例子：// find_end example
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

bool myfunction (int i, int j) {
  return (i==j);
}

int main () {
  int myints[] = {1,2,3,4,5,1,2,3,4,5};
  vector<int> myvector (myints,myints+10);
  vector<int>::iterator it;

  int match1[] = {1,2,3};

  // using default comparison:
  it = find_end (myvector.begin(), myvector.end(), match1, match1+3);

  if (it!=myvector.end())
    cout << "match1 last found at position " << int(it-myvector.begin()) << endl;

  int match2[] = {4,5,1};

  // using predicate comparison:
  it = find_end (myvector.begin(), myvector.end(), match2, match2+3, myfunction);

  if (it!=myvector.end())
    cout << "match2 last found at position " << int(it-myvector.begin()) << endl;
 
  return 0;
}
 

Output:
Match found at position 5
Match found at position 3