C++ Algorithm Library - lexicographical_compare() Function


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Description

The C++ function std::algorithm::lexicographical_compare() tests whether one range is lexicographically less than another or not. A lexicographical comparison is the kind of comparison generally used to sort words alphabetically in dictionaries.

Declaration

Following is the declaration for std::algorithm::lexicographical_compare() function form std::algorithm header.

C++98

template <class InputIterator1, class InputIterator2>
bool lexicographical_compare(InputIterator1 first1, InputIterator1 last1,
   InputIterator2 first2, InputIterator2 last2);

Parameters

  • first1 − Input iterator to the initial position of the first sequence.

  • last1 − Input iterator to the final position of the first sequence.

  • first2 − Input iterator to the initial position of the second sequence.

  • last2 − Input iterator to the final position of the second sequence.

Return value

Returns true if one range is lexicographically less than second otherwise returns false.

Exceptions

Throws exception if either the element comparison or an operation on an iterator throws exception.

Please note that invalid parameters cause undefined behavior.

Time complexity

2 * min(N1, N2) , where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).

Example

The following example shows the usage of std::algorithm::lexicographical_compare() function.

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

using namespace std;

int main(void) {
   vector<string> v1 = {"One", "Two", "Three"};
   vector<string> v2 = {"one", "two", "three"};
   bool result;

   result = lexicographical_compare(v1.begin(), v1.end(), v2.begin(), v2.end());

   if (result == true)
      cout << "v1 is less than v2." << endl;

   v1[0] = "two";

   result = lexicographical_compare(v1.begin(), v1.end(), v2.begin(), v2.end());

   if (result == false)
      cout << "v1 is not less than v2." << endl;

   return 0;
}

Let us compile and run the above program, this will produce the following result −

v1 is less than v2.
v1 is not less than v2.
algorithm.htm
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