Max Points on a Line

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

 

思想：以一个点为起点，计算以该点为起点的所有不同斜率上点的数目，从中挑选最大值。

java 代码： 注意java中int与int操作不会自动转换成double，如果结果为double，最好先转换成double。否则可能出现0.0 和 -0.0的差异。

1. public int maxPoints(Point[] points) {
2. int plen = points.length;
3. if(plen == 0) {
4. return 0;
5. }
6. int max_value = 0;
7. Map<Double,Integer> scope = new HashMap<Double,Integer>();
8. for(int i=0;i<plen;i++) {
9. int dup = 1;
10. scope.clear();
11. scope.put(Double.MAX_VALUE,0);
12. for(int j=i+1;j<plen;j++) {
13. if(points[j].y==points[i].y && points[j].x == points[i].x) {
14. dup++;
15. continue;
16. }
17. if(points[j].x == points[i].x) {
18. scope.put(Double.MAX_VALUE,scope.get(Double.MAX_VALUE) + 1);
19. continue;
20. }
21. double k = (points[j].y - points[i].y)*1.0/(points[j].x-points[i].x);  //[2,3],[3,3],[-5,5] 会得到0.0和-0.0.
22. if(k==-0.0) k=0.0;  //可以把-0.0 转换成0.0 或者直接计算时把int强转换为double
23. if(scope.containsKey(k)) {
24. scope.put(k,scope.get(k)+1);
25. } else {
26. scope.put(k,1);
27. }
28. }
29. Iterator iter = scope.entrySet().iterator();
30. while(iter.hasNext()) {
31. Map.Entry entry = (Map.Entry)iter.next();
32. Integer value = (Integer)entry.getValue();
33. if(dup + value > max_value) {
34. max_value = dup + value;
35. }
36. }
37. }
38. return max_value;
39. }

C++代码：相对简洁一些

1. int maxPoints(vector<Point> &points) {
2. int psz=points.size();
3. if(psz==0) return 0;
4. map<double,int> mp;
5. int maxpoint=0;
6. for(int i=0;i<psz;i++) {
7. mp.clear();
8. mp[INT_MIN]=0;
9. int dup=1;
10. for(int j=i+1;j<psz;j++) {
11. if(points[j].x==points[i].x && points[j].y==points[i].y) {
12. dup++;
13. } else if(points[j].x == points[i].x) {
14. mp[INT_MAX]++;
15. } else {
16. double k=(points[j].y-points[i].y)*1.0/(points[j].x-points[i].x);
17. mp[k]++;
18. }
19. }
20. for(map<double,int>::iterator it=mp.begin();it!=mp.end();it++) {
21. if(it->second + dup > maxpoint) maxpoint = it->second + dup;
22. }
23. }
24. return maxpoint;
25. }