Finding the left-most and right-most points of a list. std::find_if the right way to go?

I have a list of Point objects, (each one with x,y properties) and would like to find the left-most and right-most points. I've been trying to do it with find_if, but i'm not sure its the way to go, because i can't seem to pass a comparator instance. Is find_if the way to go? Seems not. So, is there an algorithm in <algorithm> to achieve this?

Thanks in advance.

#include <iostream> #include <list> #include <algorithm> using namespace std; typedef struct Point{ float x; float y; } Point; bool left(Point& p1,Point& p2) { return p1.x < p2.x; } int main(){ Point p1 ={-1,0}; Point p2 ={1,0}; Point p3 ={5,0}; Point p4 ={7,0}; list <Point> points; points.push_back(p1); points.push_back(p2); points.push_back(p3); points.push_back(p4); //Should return an interator to p1. find_if(points.begin(),points.end(),left); return 0; }

-------------Problems Reply------------

Use std::min_element and std::max_element instead.

list<Point>::iterator left = std::min_element(points.begin(), points.end(), left);
list<Point>::iterator right = std::max_element(points.begin(), points.end(), left);

I would also change the signature of left to:

bool left(const Point& p1, const Point& p2)

even better would be to use the boost minmax element:

http://www.boost.org/doc/libs/1_42_0/libs/algorithm/minmax/index.html

#include <boost/algorithm/minmax_element.hpp>
...
auto res = boost::minmax_element(points.begin(), points.end(), left);

std::cout << "min: " << res.first << std::endl;
std::cout << "max: " << res.second << std::endl;

If you use pair<float, float> instead of your own Point, there's no need for a special comparator. There would also be an ordering on the y-axis of points with the same x-coordinate, which can be useful.

There are various ways to imbue typedef pair<float, float> Point; with custom behavior, if you're so inclined. For example,

typedef pair<float, float> Point;

enum AxisUnit { x, y };
float &operator*( Point &p, AxisUnit c ) // "special case" of inner product
{ return c == x? p.first : p.second; }

Point my_point( 2.5, 6.3 );
float x_coord = my_point * x;

Category:c# Views:0 Time:2010-04-16

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