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:
代码语言：javascript复制输入: [[1,1],[2,2],[3,3]]
输出: 3
解释:
^
|
| o
| o
| o
------------->
0 1 2 3 4示例 2:
代码语言：javascript复制输入: [[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解：未AC，题目不难，两层for循环求斜率，对于一个固定点和另一点求得的斜率相同说明在一条直线，用例会出现重复的点，这个非常麻烦。
代码语言：javascript复制 class Point {
int x;
int y;
Point() {
x = 0;
y = 0;
}
Point(int a, int b) {
x = a;
y = b;
}
@Override
public int hashCode() {
return 10;
}
@Override
public boolean equals(Object obj) {
Point p = (Point) obj;
return this.x == p.x && this.y == p.y;
}
}
class Solution {
public int maxPoints(Point[] points) {
if (points.length == 0) {
return 0;
}
Set<Point> set = new HashSet<>(Arrays.asList(points));
if (set.size() == 1) {
return points.length;
}
int max = 0;
for (int i = 0; i < points.length; i ) {
Map<Double, Integer> map = new HashMap<>();
int yCount = 0;
int cf = 0;
for (int j = i 1; j < points.length; j ) {
if (points[i].x != points[j].x) {//不重合不垂直
double k = getK(points[i], points[j]);
if (map.containsKey(k)) {
map.put(k, map.get(k) 1);
} else {
map.put(k, 1);
}
} else if (points[i].y == points[j].y) {//重合
cf ;
} else {//垂直
yCount ;
}
}
for (Double d : map.keySet()) {
max = Math.max(max, map.get(d) cf);
}
max = Math.max(max, yCount cf);
}
return max 1;
}
private double getK(Point a, Point b) {
return (b.y - a.y) / (b.x - a.x);
}
}
