题目：
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
题解：
这道题就是给你一个2D平面，然后给你的数据结构是由横纵坐标表示的点，然后看哪条直线上的点最多。
（1）两点确定一条直线
（2）斜率相同的点落在一条直线上
（3）坐标相同的两个不同的点 算作2个点
利用HashMap，Key值存斜率，Value存此斜率下的点的个数。同时考虑特殊情况，如果恰巧遍历到一个相同坐标的点，那么就维护一个local的counter来记录相同的点。
维护一个localmax，计算当前情况下的最大值；再维护一个全局Max来计算总的最大值。
返回全局Max即可。
代码如下：
1 public int maxPoints(Point[] points) {
2 if(points.length == 0||points == null)
3 return 0;
4
5 if(points.length == 1)
6 return 1;
7
8 int max = 1; //the final max value, at least one
9 for(int i = 0; i < points.length; i++) {
10 HashMap<Float, Integer> hm = new HashMap<Float, Integer>();
11 int same = 0;
12 int localmax = 1; //the max value of current slope, at least one
13 for(int j = 0; j < points.length; j++) {
14 if(i == j)
15 continue;
16
17 if(points[i].x == points[j].x && points[i].y == points[j].y){
18 same++;
19 continue;
20 }
21
22 float slope = ((float)(points[i].y - points[j].y))/(points[i].x - points[j].x);
23
24 if(hm.containsKey(slope))
25 hm.put(slope, hm.get(slope) + 1);
26 else
27 hm.put(slope, 2); //two points form a line
28 }
29
30 for (Integer value : hm.values())
31 localmax = Math.max(localmax, value);
32
33 localmax += same;
34 max = Math.max(max, localmax);
35 }
36 return max;
37 }
2 if(points.length == 0||points == null)
3 return 0;
4
5 if(points.length == 1)
6 return 1;
7
8 int max = 1; //the final max value, at least one
9 for(int i = 0; i < points.length; i++) {
10 HashMap<Float, Integer> hm = new HashMap<Float, Integer>();
11 int same = 0;
12 int localmax = 1; //the max value of current slope, at least one
13 for(int j = 0; j < points.length; j++) {
14 if(i == j)
15 continue;
16
17 if(points[i].x == points[j].x && points[i].y == points[j].y){
18 same++;
19 continue;
20 }
21
22 float slope = ((float)(points[i].y - points[j].y))/(points[i].x - points[j].x);
23
24 if(hm.containsKey(slope))
25 hm.put(slope, hm.get(slope) + 1);
26 else
27 hm.put(slope, 2); //two points form a line
28 }
29
30 for (Integer value : hm.values())
31 localmax = Math.max(localmax, value);
32
33 localmax += same;
34 max = Math.max(max, localmax);
35 }
36 return max;
37 }
Reference：
http://blog.csdn.net/ttgump/article/details/23146357
http://blog.csdn.net/linhuanmars/article/details/21060933