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：注意可能存在重合点，除去重合点，只要记录能够表示一条直线的(x0,y0)，其中a=x1-x2和y1-y2的最大公约数，x0=(x1-x2)/a,y0=(y1-y2)/a。注意到对于特定的i0，只要第一次出现(x0,y0)，那么在直线(x0,y0)的点的数量一定可以在i0下得到最大值，本质上就是比较每个i下第一次出现的(x0,y0)包含的点的数目。

代码：

 1 /**
 2  * Definition for a point.
 3  * class Point {
 4  *     int x;
 5  *     int y;
 6  *     Point() { x = 0; y = 0; }
 7  *     Point(int a, int b) { x = a; y = b; }
 8  * }
 9  */
10 public class Solution {
11     public int maxPoints(Point[] points) {
12         HashMap<Integer, Map<Integer, Integer>> hashMap = new HashMap<Integer, Map<Integer, Integer>>();
13         int res = 0, maxCount = 0, overlap = 0;
14         for (int i = 0; i < points.length; ++i) {
15             hashMap.clear();
16             maxCount = overlap = 0;
17             for (int j = i + 1; j < points.length; j++) {
18                 int x = points[i].x - points[j].x;
19                 int y = points[i].y - points[j].y;
20                 if (x == 0 && y == 0) {//重合点
21                     overlap++;
22                     continue;
23                 }
24                 int gcd = Gcd(x, y);
25                 x = x / gcd;
26                 y = y / gcd;
27                 if (hashMap.containsKey(x)) {
28                     hashMap.get(x).put(y, hashMap.get(x).getOrDefault(y, 0) + 1);
29                 }else {
30                     Map<Integer,Integer> map = new HashMap<Integer,Integer>();
31                     map.put(y, 1);
32                     hashMap.put(x, map);
33                 }
34                 maxCount = Math.max(maxCount, hashMap.get(x).get(y));
35             }
36             res = Math.max(res, maxCount + overlap + 1);
37         }
38         return res;
39     }
40     private int Gcd(int x, int y){
41         if (y == 0) return x;
42         return Gcd(y, x%y);
43     }
44 }