CodeCogs

Searches for the last occurrence of a subrang

Definition

The find_end() algorithm is defined in the standard header <algorithm> and in the nonstandard backward-compatibility header <algo.h>.

Interface

#include <algorithm>
template < class ForwardIterator1, class ForwardIterator2 >
   ForwardIterator1 find_end(
      ForwardIterator1 first1, 
      ForwardIterator1 last1,
      ForwardIterator2 first2, 
      ForwardIterator2 last2
   );
template < class ForwardIterator1, class ForwardIterator2, class Pr >
   ForwardIterator1 find_end(
      ForwardIterator1 first1, 
      ForwardIterator1 last1,
      ForwardIterator2 first2, 
      ForwardIterator2 last2,
      BinaryPredicate comp
   );

Parameters:

Parameter	Description
first1	A forward iterator addressing the position of the first element in the range to be searched
last1	A forward iterator addressing the position one past the final element in the range to be searched
first2	A forward iterator addressing the position of the first element in the range to be searched
last2	A forward iterator addressing the position one past the final element in the range to be searched
comp	User-defined predicate function object that defines the condition to be satisfied if two elements are to be taken as equivalent. A binary predicate takes two arguments and returns true when satisfied and false when not satisfied

Description

Find_end finds the last subsequence (elements between first2 and last2) contained into a sequence (elements between first1 and last1 ).

Return Value

Returns a forward iterator addressing the position of the first element of the last subsequence that matches the specified sequence or that is equivalent in a sense specified by a binary predicate.

Complexity

The complexity is linear, find_end performs at most (last1 - first1) * (last2 - first2) comparisons.

References

http://msdn.microsoft.com/en-us/

Example 1

Problem

This program illustrates the use of the STL find_end() algorithm (default version) to find the last occurrence of one range of integer values in a vector within another range of integer values, also in a vector.

Workings

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
  int a1[] = {1, 2, 3, 4, 5, 6, 7, 8, 3, 4, 5, 9, 10, 11};
  vector<int> v1(a1, a1+14);
  cout <<"\nHere are the contents of v1:\n";
  for (vector<int>::size_type i=0; i<v1.size(); i++)
    cout <<v1.at(i)<<" ";

  int a2[] = {3, 4, 5};
  vector<int> v2(a2, a2+3);
  cout <<"\nHere are the contents of v2:\n";
  for (vector<int>::size_type i=0; i<v2.size(); i++)
    cout <<v2.at(i)<<" ";

  vector<int>::iterator p;
  
  p = find_end(v1.begin(), v1.end(), v2.begin(), v2.end());
  if (p != v1.end())
    cout <<"\nThe last instance of v2 in v1 begins at location "
         <<(int)(p-v1.begin()+1)<<".";
  else
    cout <<"\nNo instance of v2 was found in v1.";

  return 0;
}

Solution

Output:

Here are the contents of v1:
1 2 3 4 5 6 7 8 3 4 5 9 10 11

Here are the contents of v2:
3 4 5

The last instance of v2 in v1 begins at location 9.

References

http://www.smu.ca/academic/science/compsci/

Example 2

Problem

This program illustrates the use of the STL find_end() algorithm (extended version) to find the last occurrence of one range of matching integer values in a vector within another range of integer values, also in a vector, and where corresponding values match if and only if they have the same digit sum.

Workings

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// Tests if two positive integers have the same digit sum.

bool digitSumsAreSame
(
  int i1, //in
  int i2 //in
)

{
  int digitSum1 = 0;
  while (i1 != 0)
  {
    digitSum1 += i1 % 10;
    i1 /= 10;
  }

  int digitSum2 = 0;
  while (i2 != 0)
  {
    digitSum2 += i2 % 10;
    i2 /= 10;
  }
  return digitSum1 == digitSum2;
}

int main()
{
  int a1[] = {17, 21, 13, 5, 6, 7, 8, 12, 31, 14, 9, 10, 11};
  vector<int> v1(a1, a1+13);
  cout <<"\nHere are the contents of v1:\n";
  for (vector<int>::size_type i=0; i<v1.size(); i++)
    cout << v1.at(i) << " ";

  int a2[] = {21, 13, 5};
  vector<int> v2(a2, a2+3);
  cout <<"\nHere are the contents of v2:\n";
  for (vector<int>::size_type i=0; i<v2.size(); i++)
    cout <<v2.at(i)<<" ";

  vector<int>::iterator p;
  
  p = find_end(v1.begin(), v1.end(), v2.begin(), v2.end(),digitSumsAreSame);
  if (p != v1.end())
    cout <<"\nThe last instance of v2 in v1 begins at location "
         <<(int)(p-v1.begin()+1)<<".";
  else
    cout <<"\nNo instance of v2 was found in v1.";

  return 0;
}

Solution

Output:

Here are the contents of v1:
17 21 13 5 6 7 8 12 31 14 9 10 11

Here are the contents of v2:
21 13 5

The last instance of v2 in v1 begins at location 8.

References

http://www.smu.ca/academic/science/compsci/

find_end

Definition

Interface

Description

Return Value

Complexity

References

Example 1

Example 2

See Also