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
public class MaxPointsonaLine {
// 已经给定的Point class
class Point {
int x;
int y;
Point() {
x = 0;
y = 0;
}
Point(int a, int b) {
x = a;
y = b;
}
}
public int maxPoints(Point[] points) {
int result = 0;
Map<Integer, Map<Integer, Integer>> map = new HashMap<>();
// standing at points[i]
for (int i = 0; i < points.length; i++) {
map.clear();
int overlapping = 0;
int nonoverlapping = 0;
// checking points[j]
for (int j = i + 1; j < points.length; j++) {
int x = points[j].x - points[i].x;
int y = points[j].y - points[i].y;
if (x == 0 && y == 0) {
overlapping++;
continue;
}
int gcd = generateGCD(x, y);
if (gcd != 0) {
x = x / gcd;
y = y / gcd;
}
if (map.containsKey(x)) {
if (map.get(x).containsKey(y)) {
map.get(x).put(y, map.get(x).get(y) + 1);
} else {
map.get(x).put(y, 1);
}
} else {
Map<Integer, Integer> m = new HashMap<>();
m.put(y, 1);
map.put(x, m);
}
overlapping = Math.max(nonoverlapping, map.get(x).get(y));
}
result = Math.max(result, overlapping + nonoverlapping + 1);
}
return result;
}
public int generateGCD(int a, int b) {
return (b == 0) ? a : generateGCD(b, a % b);
}
}