Max Points on a Line(Hard)

Leetcode Source

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

任意一条直线都可以表述为

y = ax + b

假设，有两个点(x1,y1), (x2,y2)，如果他们都在这条直线上则有

y1 = kx1 +b

y2 = kx2 +b

由此可以得到关系，k = (y2-y1)/(x2-x1)。即如果点c和点a的斜率为k, 而点b和点a的斜率也为k，可以知道点c和点b也在一条线上。

取定一个点points[i], 遍历其他所有节点, 然后统计斜率相同的点数，并求取最大值即可

/**
 * 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) {
        int length = points.length;
        if (length < 3)
            return length;
        int max = 2;
        for (int i = 0; i < length; i++) {
            int pointMax = 1, samePointCount = 0;
            HashMap<Double, Integer> slopeCount = new HashMap<Double, Integer>();
            Point origin = points[i];
            for (int j = i + 1; j < length; j++) {
                Point target = points[j];
                if (origin.x == target.x && origin.y == target.y) {
                    samePointCount++;
                    continue;
                }
                double k;
                if (origin.x == target.x) {
                    k = Float.POSITIVE_INFINITY;
                } else if (origin.y == target.y) {
                    k = 0;
                } else {
                    k = ((float) (origin.y  -target.y)) / (origin.x - target.x);
                }
                if (slopeCount.containsKey(k)) {
                    slopeCount.put(k, slopeCount.get(k) + 1);
                } else {
                    slopeCount.put(k, 2);
                }
                pointMax = Math.max(pointMax, slopeCount.get(k));
            }
            pointMax += samePointCount;
            max = Math.max(pointMax, max);
        }
        return max;
    }
}

References

1. from SegmentFault