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.

思路： 从 points[] 中依次取出一个点，然后依次和后面一个点组成一条直线。 y = ax+b;

然后再遍历所有的点看是否满足当前的斜率a 和 截距 b。 如果满足count++。维护一个max， 每次遍历完将count和max比较取最大.

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    public int maxPoints(Point[] points) {
        if(points == null || points.length == 0)
            return 0;
        int len = points.length;
        int max = 1;
        for(int i = 0; i < len; i++){
            Point point1 = points[i];
            for(int j = i+1; j<len; j++){
                Point point2 = points[j];
                int count = 0;
                
                if(point1.x == point2.x){
                    for(int z = 0; z < len; z++){
                        if(points[z].x == point1.x)
                            count++;
                    }
                }else{
                       double x1 = point1.x;
                        double y1 = point1.y;
                        double x2 = point2.x;
                        double y2 = point2.y;
                        double a = (y1 - y2) / (x1 - x2);
                        // 这里的b 是根据 y1 = ax1+b 和 y2 = ax2+b 求出来的
                        double b = (y1 * x2 - y2 * x1) / (x2 - x1);

                    for(int z = 0; z < len; z++){
                        if(Math.abs(points[z].x * a + b - points[z].y) < 0.000001)
                            count++;
                    }
                }

                if(count > max)
                    max = count;
            }
        }
        return max;
    }
}