Finding local extremum in vector of numbers [duplicate] - c++

Question

How can I get the maximum or minimum value in a vector in C++?
And am I wrong in assuming it would be more or less the same with an array?
I need an iterator, right? I tried it with max_element, but I kept getting an error.
vector<int>::const_iterator it;
it = max_element(cloud.begin(), cloud.end());
error: request for member ‘begin’ in ‘cloud’, which is of non-class type ‘int [10]’

Answer 1

Using C++11/C++0x compile flags, you can
auto it = max_element(std::begin(cloud), std::end(cloud)); // C++11
Otherwise, write your own:
template <typename T, size_t N> const T* mybegin(const T (&a)[N]) { return a; }
template <typename T, size_t N> const T* myend (const T (&a)[N]) { return a+N; }
See it live at http://ideone.com/aDkhW:
#include <iostream>
#include <algorithm>
template <typename T, size_t N> const T* mybegin(const T (&a)[N]) { return a; }
template <typename T, size_t N> const T* myend (const T (&a)[N]) { return a+N; }
int main()
{
const int cloud[] = { 1,2,3,4,-7,999,5,6 };
std::cout << *std::max_element(mybegin(cloud), myend(cloud)) << '\n';
std::cout << *std::min_element(mybegin(cloud), myend(cloud)) << '\n';
}
Oh, and use std::minmax_element(...) if you need both at once :/

Answer 2

If you want to use the function std::max_element(), the way you have to do it is:
double max = *max_element(vector.begin(), vector.end());
cout<<"Max value: "<<max<<endl;

Answer 3

Let,
#include <vector>
vector<int> v {1, 2, 3, -1, -2, -3};
If the vector is sorted in ascending or descending order then you can find it with complexity O(1).
For a vector of ascending order the first element is the smallest element, you can get it by v[0] (0 based indexing) and last element is the largest element, you can get it by v[sizeOfVector-1].
If the vector is sorted in descending order then the last element is the smallest element,you can get it by v[sizeOfVector-1] and first element is the largest element, you can get it by v[0].
If the vector is not sorted then you have to iterate over the vector to get the smallest/largest element.In this case time complexity is O(n), here n is the size of vector.
int smallest_element = v[0]; //let, first element is the smallest one
int largest_element = v[0]; //also let, first element is the biggest one
for(int i = 1; i < v.size(); i++) //start iterating from the second element
{
if(v[i] < smallest_element)
{
smallest_element = v[i];
}
if(v[i] > largest_element)
{
largest_element = v[i];
}
}
You can use iterator,
for (vector<int>:: iterator it = v.begin(); it != v.end(); it++)
{
if(*it < smallest_element) //used *it (with asterisk), because it's an iterator
{
smallest_element = *it;
}
if(*it > largest_element)
{
largest_element = *it;
}
}
You can calculate it in input section (when you have to find smallest or largest element from a given vector)
int smallest_element, largest_element, value;
vector <int> v;
int n;//n is the number of elements to enter
cin >> n;
for(int i = 0;i<n;i++)
{
cin>>value;
if(i==0)
{
smallest_element= value; //smallest_element=v[0];
largest_element= value; //also, largest_element = v[0]
}
if(value<smallest_element and i>0)
{
smallest_element = value;
}
if(value>largest_element and i>0)
{
largest_element = value;
}
v.push_back(value);
}
Also you can get smallest/largest element by built in functions
#include<algorithm>
int smallest_element = *min_element(v.begin(),v.end());
int largest_element = *max_element(v.begin(),v.end());
You can get smallest/largest element of any range by using this functions. such as,
vector<int> v {1,2,3,-1,-2,-3};
cout << *min_element(v.begin(), v.begin() + 3); //this will print 1,smallest element of first three elements
cout << *max_element(v.begin(), v.begin() + 3); //largest element of first three elements
cout << *min_element(v.begin() + 2, v.begin() + 5); // -2, smallest element between third and fifth element (inclusive)
cout << *max_element(v.begin() + 2, v.begin()+5); //largest element between third and first element (inclusive)
I have used asterisk (*), before min_element()/max_element() functions. Because both of them return iterator. All codes are in c++.

Answer 4

You can print it directly using the max_element or min_element function.
For example:
cout << *max_element(v.begin(), v.end());
cout << *min_element(v.begin(), v.end());

Answer 5

Assuming cloud is int cloud[10] you can do it like this:
int *p = max_element(cloud, cloud + 10);

Answer 6

In C++11, you can use some function like that:
int maxAt(std::vector<int>& vector_name) {
int max = INT_MIN;
for (auto val : vector_name) {
if (max < val) max = val;
}
return max;
}

Answer 7

If you want to use an iterator, you can do a placement-new with an array.
std::array<int, 10> icloud = new (cloud) std::array<int,10>;
Note the lack of a () at the end, that is important. This creates an array class that uses that memory as its storage, and has STL features like iterators.
(This is C++ TR1/C++11 by the way)

Answer 8

You can use max_element to get the maximum value in vector.
The max_element returns an iterator to largest value in the range, or last if the range is empty. As an iterator is like pointers (or you can say pointer is a form of iterator), you can use a * before it to get the value.
So as per the problem you can get the maximum element in an vector as:
int max=*max_element(cloud.begin(), cloud.end());
It will give you the maximum element in your vector "cloud".
Hope it helps.

Answer 9

Answer on the behalf of the author
for (unsigned int i = 0; i < cdf.size(); i++)
if (cdf[i] < cdfMin)
cdfMin = cdf[i];
where cdf is a vector.

Answer 10

Just this:
// assuming "cloud" is:
// int cloud[10];
// or any other fixed size
#define countof(x) (sizeof(x)/sizeof((x)[0]))
int* pMax = std::max_element(cloud, cloud + countof(cloud));