Leetcode 149. 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.

思路是：用两个循环去查看所有情况

same 表示两个点完全重合

sameX 表示两个点有相同的x坐标

public int maxPoints(Point[] points) {        if (points == null || points.length == 0) return 0;        if (points.length <= 2) return points.length;                int res = 0, len = points.length;        for (int i = 0; i < len; i++) {            Point start = points[i];            int same = 1, sameX = 0, max = 0;            Map<String, Integer> hm = new HashMap<>();            for (int j = 0; j < len; j++) {                if (i == j) continue;                Point end = points[j];                if (start.x == end.x && start.y == end.y) same++;                else if (start.x == end.x) sameX++;                else {                    String slope = getSlope(start, end);                    hm.put(slope, hm.getOrDefault(slope, 0) + 1);                    max = Math.max(max, hm.get(slope)); 计算出相同斜率的最大的个数                }            }            max = Math.max(max, sameX) + same;  从相同斜率和有相同X坐标的点选出最大的 加上 点完全重合点的个数            res = Math.max(res, max);        }        return res;    }        private String getSlope(Point p1, Point p2) {        int a = p1.y - p2.y;        int b = p1.x - p2.x;        if (b == 0) return null;        int gcd = a >= b ? gcd(a, b) : gcd(b, a);        a /= gcd;        b /= gcd;        return a + "/" + b;    }        private int gcd(int a, int b) {        if (b == 0) return a;        return gcd(b, a % b);    }