Max Points on a Line

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

/*    *  A line is determined by two factors,say y=ax+b    *      *  If two points(x1,y1) (x2,y2) are on the same line(Of course).     *  Consider the gap between two points.    *  We have (y2-y1)=a(x2-x1),a=(y2-y1)/(x2-x1) a is a rational, b is canceled since b is a constant    *  If a third point (x3,y3) are on the same line. So we must have y3=ax3+b    *  Thus,(y3-y1)/(x3-x1)=(y2-y1)/(x2-x1)=a    *  Since a is a rational, there exists y0 and x0, y0/x0=(y3-y1)/(x3-x1)=(y2-y1)/(x2-x1)=a    *  So we can use y0 and x0 to track a line;    */public class Solution {public int maxPoints(Point[] points) {if (points == null)return 0;if (points.length <= 2)return points.length;Map<Integer, Map<Integer, Integer>> map = new HashMap<Integer, Map<Integer, Integer>>();int result = 0;for (int i = 0; i < points.length; i++) {map.clear();int overlap = 0, max = 0;for (int j = i + 1; j < points.length; j++) {int x = points[j].x - points[i].x;int y = points[j].y - points[i].y;//重复的点if (x == 0 && y == 0) {overlap++;continue;}int gcd = generateGCD(x, y);if (gcd != 0) {x /= gcd;y /= gcd;}if (map.containsKey(x)) {if (map.get(x).containsKey(y)) {map.get(x).put(y, map.get(x).get(y) + 1);} else {map.get(x).put(y, 1);}} else {Map<Integer, Integer> m = new HashMap<Integer, Integer>();m.put(y, 1);map.put(x, m);}max = Math.max(max, map.get(x).get(y));}result = Math.max(result, max + overlap + 1);}return result;}//g(64,3)//g(3,1)//g(1,0)//return 1//g(64,16)//g(16,0)//return 16//求最大公约数private int generateGCD(int a, int b) {if (b == 0)return a;elsereturn generateGCD(b, a % b);}public static void main(String[] args) {Solution solution = new Solution();Point p1 = new Point(1, 1);Point p2 = new Point(2, 2);Point p3 = new Point(3, 3);Point p4 = new Point(4, 4);Point[] points = {p1,p2,p3,p4};int res = solution.maxPoints(points);System.out.println(res);}}

GCD:

http://www.cppblog.com/CodePanada/archive/2011/05/19/146743.html