Leetcode 149.直线上最多的点数

直线上最多的点数

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 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

 

 1 class Solution {
 2     public int maxPoints(Point[] points) {
 3         if(points.length == 0) {
 4             return 0;
 5         }
 6         int[] count = new int[points.length];
 7         for (int i = 0; i < points.length; i++) {
 8             count[i] = 1;
 9             int size = 1;
10             int same = 0;
11             HashMap<Integer[], Integer> hashMap = new HashMap<>();
12             for (int j = 0; j < points.length; j++) {
13                 if(i != j) {
14                     if(points[i].x != points[j].x) {
15                         int dy = points[i].y - points[j].y;
16                         int dx = points[i].x - points[j].x;
17                         int gcd = generateGCD(dy, dx);
18                         if(gcd != 0) {
19                             dy = dy / gcd;
20                             dx = dx / gcd;
21                         }
22                         Integer[] nums = new Integer[2];
23                         nums[0] = dy;
24                         nums[1] = dx;
25                         boolean flag = false;
26                         for (Integer[] array : hashMap.keySet()) {
27                             if(nums[0] == array[0] && nums[1] == array[1]) {
28                                 flag = true;
29                                 hashMap.put(array, hashMap.get(array) + 1);
30                             }
31                         }
32                         if(!flag) {
33                             hashMap.put(nums, 1);
34                         }
35                     }else {
36                         if(points[i].y == points[j].y) {
37                             same++;
38                         }
39                         size++;
40                     }
41                 }
42             }
43             for (Integer[] array : hashMap.keySet()) {
44                 if(hashMap.get(array) + 1 > count[i]) {
45                     count[i] = hashMap.get(array) + 1;
46                 }
47             }
48             count[i] += same;
49             count[i] = Math.max(count[i], size);
50         }
51         int maxIndex = 0;
52         for (int i = 1; i < count.length; i++) {
53             if(count[i] > count[maxIndex]) {
54                 maxIndex = i;
55             }
56         }
57         return count[maxIndex];
58     }
59 
60     // 欧几里得算法：计算最大公约数
61     private int generateGCD(int x, int y) {
62         if (y == 0)
63             return x;
64         return generateGCD(y, x % y);
65     }
66 }

 

posted on 2018-12-29 14:51  kexinxin  阅读(126)  评论(0)  编辑  收藏  举报

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