[leetcode] 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.

https://oj.leetcode.com/problems/max-points-on-a-line/

思路：以每一个点位轴点，计算其他点的斜率，如果斜率相同则共线，用hashmap记录共线点数，不断更新最大值。

　　注意：有重复点需要处理，尤其是所有点都是重复点的情况需要特殊处理。

import java.util.HashMap;
import java.util.Map;

/**
 * http://blog.csdn.net/xhu_eternalcc/article/details/24556553
 * http://www.cnblogs.com/TenosDoIt/p/3444086.html
 * @author jd
 *
 */
public class Solution {
    public int maxPoints(Point[] points) {
        if (points == null)
            return 0;
        int n = points.length;
        if (n <= 2)
            return n;
        Map<Double, Integer> map = new HashMap<Double, Integer>();
        int maxN = 0;

        for (int i = 0; i < points.length; i++) {
            map.clear();
            Point cur = points[i];
            int duplicates = 1;

            boolean notAllDup = false;
            for (int j = 0; j < points.length; j++) {
                if (j == i)
                    continue;
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    duplicates++;
                } else {
                    notAllDup = true;
                    Double slope = slope(cur, points[j]);
                    if (map.get(slope) == null) {
                        map.put(slope, 1);
                    } else {
                        map.put(slope, map.get(slope) + 1);
                    }
                }
            }

            if (!notAllDup) {
                maxN = Math.max(maxN, duplicates);
            }

            for (Double key : map.keySet()) {
                maxN = Math.max(maxN, map.get(key) + duplicates);
            }

        }

        return maxN;
    }

    private double slope(Point cur, Point p) {
        int x1 = cur.x;
        int y1 = cur.y;
        int x2 = p.x;
        int y2 = p.y;
        if (x1 == x2 && y1 == y2) {
            return -Double.MAX_VALUE;
        } else if (x1 == x2)
            return Double.MAX_VALUE;
        else if (y1 == y2)
            return 0;
        else
            return 1.0 * (y2 - y1) / (x2 - x1);

    }

    public static void main(String[] args) {
        Point[] points = null;
        // 3
        points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(9, 10), new Point(10, 9) };
        System.out.println(new Solution().maxPoints(points));

        // 5
        points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(10, 10), new Point(9, 9) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(0, 0), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 0) };
        System.out.println(new Solution().maxPoints(points));

        // 1
        points = new Point[] { new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 2
        points = new Point[] { new Point(1, 1), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 2
        points = new Point[] { new Point(0, 0), new Point(1, 1), new Point(1, -1) };
        System.out.println(new Solution().maxPoints(points));
    }

}

第二遍记录：

　　对于求slope，注意斜率无限大的情况和两个点重合的情况。

　　统计点时，注意与当前点的重合点的统计，以及所有点都是重合点的特殊情况。

第三遍记录：

　　重新写了一边，对于每个节点依次统计跟其他节点的斜率数， 注意跟当前节点重合的点要分开统计。

import java.util.HashMap;

public class Solution {

    public int maxPoints(Point[] points) {
        if (points == null)
            return 0;
        if (points.length <= 2)
            return points.length;

        int max = 2;
        // at least 3 points now
        for (int i = 0; i < points.length; i++) {
            Point cur = points[i];
            HashMap<Double, Integer> map = new HashMap<Double, Integer>();
            int duplicates = 1;
            int curMax = 0;
            for (int j = 0; j < points.length; j++) {
                if (i == j)
                    continue;
                if (cur.x == points[j].x && cur.y == points[j].y) {
                    duplicates++;
                } else {
                    double curSlope = slope(cur, points[j]);
                    if (map.containsKey(curSlope)) {
                        map.put(curSlope, map.get(curSlope) + 1);
                    } else {
                        map.put(curSlope, 1);
                    }
                }

            }

            for (Double each : map.keySet()) {
                curMax = Math.max(curMax, map.get(each));
            }

            max = Math.max(max, duplicates + curMax);

        }

        return max;
    }

    private double slope(Point a, Point b) {
        int x1 = a.x;
        int y1 = a.y;
        int x2 = b.x;
        int y2 = b.y;
        if (x1 == x2 && y1 == y2) {
            return Integer.MIN_VALUE;
        } else if (y1 == y2) {
            return 0;
        } else if (x1 == x2) {
            return Integer.MAX_VALUE;
        } else {
            return 1.0 * (y1 - y2) / (x1 - x2);
        }

    }

    public static void main(String[] args) {
        Point[] points = null;
        // 3
        points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(9, 10), new Point(10, 9) };
        System.out.println(new Solution().maxPoints(points));

        // 5
        points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(10, 10), new Point(9, 9) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(0, 0), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 3
        points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 0) };
        System.out.println(new Solution().maxPoints(points));

        // 1
        points = new Point[] { new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 2
        points = new Point[] { new Point(1, 1), new Point(1, 1) };
        System.out.println(new Solution().maxPoints(points));

        // 2
        points = new Point[] { new Point(0, 0), new Point(1, 1), new Point(1, -1) };
        System.out.println(new Solution().maxPoints(points));
    }

}

class Point {
    int x;
    int y;

    public Point(int x, int y) {
        this.x = x;
        this.y = y;
    }

}

参考：

http://blog.csdn.net/xhu_eternalcc/article/details/24556553

http://www.cnblogs.com/TenosDoIt/p/3444086.html