[LeetCode] 149. 直线上最多的点数

题目链接 : https://leetcode-cn.com/problems/max-points-on-a-line/

题目描述:

给定一个二维平面，平面上有 n 个点，求最多有多少个点在同一条直线上。

示例:

示例 1:

输入: [[1,1],[2,2],[3,3]]
输出: 3
解释:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

示例 2:

输入: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
输出: 4
解释:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

思路:

一句话解释: 固定一点, 找其他点和这个点组成直线, 统计他们的斜率!

这里有一个重点: 求斜率.用两种方法

1. 用最大约数方法(gcd), 把他化成最简形式, 3/6 == 2/4 == 1/2
2. 除数(不太精确, 速度快!)

直接写在代码里了!

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        from collections import Counter, defaultdict
        # 所有点统计
        points_dict = Counter(tuple(point) for point in points)
        # 把唯一点列举出来
        not_repeat_points = list(points_dict.keys())
        n = len(not_repeat_points)
        if n == 1: return points_dict[not_repeat_points[0]]
        res = 0
        # 求最大公约数
        def gcd(x, y):
            if y == 0:
                return x
            else:
                return gcd(y, x % y)

        for i in range(n - 1):
            # 点1
            x1, y1 = not_repeat_points[i][0], not_repeat_points[i][1]
            # 斜率
            slope = defaultdict(int)
            for j in range(i + 1, n):
                # 点2
                x2, y2 = not_repeat_points[j][0], not_repeat_points[j][1]
                dy, dx = y2 - y1, x2 - x1
                # 方式一 利用公约数
                g = gcd(dy, dx)
                if g != 0:
                    dy //= g
                    dx //= g
                slope["{}/{}".format(dy, dx)] += points_dict[not_repeat_points[j]]
                # --------------------
                # 方式二, 利用除法(不准确, 速度快)
                # if dx == 0:
                #     tmp = "#"
                # else:
                #     tmp = dy * 1000 / dx * 1000
                # slope[tmp] += points_dict[not_repeat_points[j]]
                #------------------------------
            res = max(res, max(slope.values()) + points_dict[not_repeat_points[i]])
        return res

java

class Solution {
    public int maxPoints(int[][] points) {
        int n = points.length;
        if (n == 0) return 0;
        if (n == 1) return 1;
        int res = 0;
        for (int i = 0; i < n - 1; i++) {
            Map<String, Integer> slope = new HashMap<>();
            int repeat = 0;
            int tmp_max = 0;
            for (int j = i + 1; j < n; j++) {
                int dy = points[i][1] - points[j][1];
                int dx = points[i][0] - points[j][0];
                if (dy == 0 && dx == 0) {
                    repeat++;
                    continue;
                }
                int g = gcd(dy, dx);
                if (g != 0) {
                    dy /= g;
                    dx /= g;
                }
                String tmp = String.valueOf(dy) + "/" + String.valueOf(dx);
                slope.put(tmp, slope.getOrDefault(tmp, 0) + 1);
                tmp_max = Math.max(tmp_max, slope.get(tmp));
            }
            res = Math.max(res, repeat + tmp_max + 1);
        }
        return res;
    }

    private int gcd(int dy, int dx) {
        if (dx == 0) return dy;
        else return gcd(dx, dy % dx);
    }
}