STL algorithm minmax, minmax_element (36), stlalgorithm

STL algorithm minmax, minmax_element (36), stlalgorithm

Minmax prototype:

Std: minmax

• C ++ 11
• C ++ 14

Default (1)	template <class T> pair <const T&,const T&> minmax (const T& a, const T& b);
Custom (2)	template <class T, class Compare> pair <const T&,const T&> minmax (const T& a, const T& b, Compare comp);
Initializer list (3)	template <class T> pair<T,T> minmax (initializer_list<T> il);template <class T, class Compare> pair<T,T> minmax (initializer_list<T> il, Compare comp);

This function returns a pair. The value of the first element of the pair is the minimum value in a and B, and the value of second is the maximum value in a. B.

Use operator <for comparison.

If a = B, make_pair (a, B) is returned ).

For (3), return the pair of the minimum and maximum values in the initialization list. If the maximum value is more than one, firstf returns the minimum value of the first occurrence, and second returns the maximum value of the last occurrence.

The behavior is similar:

template <class T> pair <const T&,const T&> minmax (const T& a, const T& b) {  return (b<a) ? std::make_pair(b,a) : std::make_pair(a,b);}

A simple example:

#include <iostream>#include <algorithm>#include <vector>using namespace std;void minmax2(){    auto it=minmax(20,10);   cout<<" auto it=minmax(20,10)"<<endl;   cout<<"it.first()="<<it.first<<" ,it.second="<<it.second<<endl;   auto it2=minmax({2,3,5,7,1,3,1});   cout<<"auto it2=minmax({2,3,5,7,1,3,1})"<<endl;   cout<<"it2.first()="<<it2.first<<" ,it2.second="<<it2.second<<endl;}

Run:

Minmax_element prototype:

Std: minmax_element

Default (1)	template <class ForwardIterator> pair<ForwardIterator,ForwardIterator> minmax_element (ForwardIterator first, ForwardIterator last);
Custom (2)	template <class ForwardIterator, class Compare> pair<ForwardIterator,ForwardIterator> minmax_element (ForwardIterator first, ForwardIterator last, Compare comp);

This function is a pair composed of elements that return the maximum and minimum values in a specified range. If the maximum value is more than one, firstf returns the first iterator with the minimum value, second returns the last iterator with the maximum value.

Use operator <for comparison.

A simple example:

#include <iostream>#include <algorithm>#include <vector>using namespace std;void minmaxelement(){    vector<int> vi{3,5,4,1,3,1,9,9,5};    cout<<"vi=";    for(int i:vi)        cout<<i<<" ";    cout<<endl;    auto it=minmax_element(vi.begin(),vi.end());   cout<<" auto it=minmax_element(vi.begin(),vi.end())"<<endl;   cout<<"*it.first="<<*it.first<<" ,*it.second="<<*it.second<<endl;    cout<<"*(it.first-1)="<<*(it.first-1)<<" ,*(it.second-1)="<<*(it.second-1)<<endl;}

Run: we can see that firstf returns the first iterator with the minimum value and second returns the last iterator with the maximum value.

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1. Write a program and use MINMAX to solve the minimum and maximum values of the array, and test the array with the actual array. 2. In the algorithm

# Include <stdio. h>
Void minmax (int array [], int begin, int end, int & min, int & max, int & count)
{
Int mid = (begin + end)/2;
If (begin = end)
{
Count ++;
If (array [mid] <min ){
Min = array [mid];
} Else {
If (array [mid]> max)
Max = array [mid];
Count ++;
}
Return;
}
Minmax (array, begin, mid, min, max, count );
Minmax (array, mid + 1, end, min, max, count );
}

Void main ()
{
Int array [10] = {9, 8, 7, 6, 5, 4, 3, 2 };
Int min = array [0], max =-1, count = 0;
Minmax (array, 0, 9, min, max, count );
Printf ("min = % d, max = % d, count = % d \ n", min, max, count );
}