[LeetCode] 149. Max Points on a Line 共线点个数
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
给一个由n个点组成的2D平面，找出最多的同在一条直线上的点的个数。
共线点的条件是斜率一样，corn case：点相同；x坐标相同。
Java：
public class Solution {
public int maxPoints(Point[] points) {
int res = 0;
for (int i = 0; i < points.length; ++i) {
Map<Map<Integer, Integer>, Integer> m = new HashMap<>();
int duplicate = 1;
for (int j = i + 1; j < points.length; ++j) {
if (points[i].x == points[j].x && points[i].y == points[j].y) {
++duplicate; continue;
}
int dx = points[j].x - points[i].x;
int dy = points[j].y - points[i].y;
int d = gcd(dx, dy);
Map<Integer, Integer> t = new HashMap<>();
t.put(dx / d, dy / d);
m.put(t, m.getOrDefault(t, 0) + 1);
}
res = Math.max(res, duplicate);
for (Map.Entry<Map<Integer, Integer>, Integer> e : m.entrySet()) {
res = Math.max(res, e.getValue() + duplicate);
}
}
return res;
}
public int gcd(int a, int b) {
return (b == 0) ? a : gcd(b, a % b);
}
}
Python：
class Point:
def __init__(self, a=0, b=0):
self.x = a
self.y = b
class Solution(object):
def maxPoints(self, points):
"""
:type points: List[Point]
:rtype: int
"""
max_points = 0
for i, start in enumerate(points):
slope_count, same = collections.defaultdict(int), 1
for j in xrange(i + 1, len(points)):
end = points[j]
if start.x == end.x and start.y == end.y:
same += 1
else:
slope = float("inf")
if start.x - end.x != 0:
slope = (start.y - end.y) * 1.0 / (start.x - end.x)
slope_count[slope] += 1
current_max = same
for slope in slope_count:
current_max = max(current_max, slope_count[slope] + same)
max_points = max(max_points, current_max)
return max_points
C++：
class Solution {
public:
int maxPoints(vector<Point>& points) {
int res = 0;
for (int i = 0; i < points.size(); ++i) {
map<pair<int, int>, int> m;
int duplicate = 1;
for (int j = i + 1; j < points.size(); ++j) {
if (points[i].x == points[j].x && points[i].y == points[j].y) {
++duplicate; continue;
}
int dx = points[j].x - points[i].x;
int dy = points[j].y - points[i].y;
int d = gcd(dx, dy);
++m[{dx / d, dy / d}];
}
res = max(res, duplicate);
for (auto it = m.begin(); it != m.end(); ++it) {
res = max(res, it->second + duplicate);
}
}
return res;
}
int gcd(int a, int b) {
return (b == 0) ? a : gcd(b, a % b);
}
};
