程序媛詹妮弗
终身学习

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

Example 1:

Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

Example 2:

Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

 

题意:

给定二维平面上一些点,问最多多少个点共线

 

Solution1: HashMap

解本题需要的背景知识:【Math Fact】All the points in a line share the same slop.

                                   

The question is like standing at points[i],  find max number of points in points[j], such that points[i] and points[j] are on the same line.

1. If points[i], points[j] 's coordinator are the same, they are overlapping.

2. Otherwise, they are nonoverlapping. Based on the fact that "All the points in a line share the same slop", we use the greatest common divisor(最大公约数) to get the lowest term(最简化) for points[i], points[j]'s coordinator.  即[2,4] 和[4,8], 我们用求最大公约数的方式,将其斜率化成最简形式: 1/2 和 1/2 

3. We use Map<x, Map<y, occurance>> map to get such slop from x and y's occurance. Then we know how many non-overlapping points in such line.

 

 

code

 1 public class MaxPointsonaLine {
 2     // 已经给定的Point class
 3     class Point {
 4         int x;
 5         int y;
 6 
 7         Point() {
 8             x = 0;
 9             y = 0;
10         }
11 
12         Point(int a, int b) {
13             x = a;
14             y = b;
15         }
16     }
17 
18     public int maxPoints(Point[] points) {
19         int result = 0;
20         Map<Integer, Map<Integer, Integer>> map = new HashMap<>();
21         // standing at points[i]
22         for (int i = 0; i < points.length; i++) {
23             map.clear();
24             int overlapping = 0;
25             int nonoverlapping = 0;
26             // checking points[j] 
27             for (int j = i + 1; j < points.length; j++) {
28                 int x = points[j].x - points[i].x;
29                 int y = points[j].y - points[i].y;
30                 if (x == 0 && y == 0) {
31                     overlapping++;
32                     continue;
33                 }
34                 int gcd = generateGCD(x, y);
35                 if (gcd != 0) {
36                     x = x / gcd;
37                     y = y / gcd;
38                 }
39                 if (map.containsKey(x)) {
40                     if (map.get(x).containsKey(y)) {
41                         map.get(x).put(y, map.get(x).get(y) + 1);
42                     } else {
43                         map.get(x).put(y, 1);
44                     }
45                 } else {
46                     Map<Integer, Integer> m = new HashMap<>();
47                     m.put(y, 1);
48                     map.put(x, m);
49                 }
50                 overlapping = Math.max(nonoverlapping, map.get(x).get(y));
51             }
52             result = Math.max(result, overlapping + nonoverlapping + 1);
53         }
54         return result;
55     }
56 
57     public int generateGCD(int a, int b) {
58         return (b == 0) ? a : generateGCD(b, a % b);
59     }
60 }

 

posted on 2019-04-25 21:58  程序媛詹妮弗  阅读(297)  评论(0编辑  收藏  举报