Princeton Algorithm Part 1 Week

时间给的非常充裕(N^4)，而且是穷举的方式 比较简单。 下一个 Fast class 就相对难一些了

Brute Collinear Points:

public class BruteCollinearPoints {
    private LineSegment[] segarr = new LineSegment[1];
    private int           segnum = 0;

    public BruteCollinearPoints(Point[] points) {
        // finds all line segments containing 4 points
        int n = points.length;
        for (int p = 0; p < n; p++) {
            for (int q = p + 1; q < n; q++) {
                for (int r = q + 1; r < n; r++)
                    if (points[p].slopeTo(points[q]) 
                            == points[p].slopeTo(points[r])) {
                        for (int s = r + 1; s < n; s++) {
                            if (points[p].slopeTo(points[q]) 
                                    == points[p].slopeTo(points[s])) {
                                segarrin(new LineSegment
                                        (points[p], points[s])); //??first try
                            }
                        }
                    }
            }
        }
    }

    private void segarrin(LineSegment line) {
        // Debugged;
        if (segarr.length == segnum) {
            LineSegment[] temp = segarr;
            segarr = new LineSegment[(segnum + 1) * 2];
            for (int i = 0; i < segnum; i++) {
                segarr[i] = temp[i];
            }
        }
        segarr[segnum++] = line;
    }

    public int numberOfSegments() {
        return segnum;
    }

    public LineSegment[] segments() {
        return segarr;
    }

    private void printer() {
        for (int i = 0; i < segnum; i++) {
            segarr[i].draw();
            System.out.println(segarr[i].toString());
        }
    }
}