1 /* 2 * A line is determined by two factors,say y=ax+b 3 * 4 * If two points(x1,y1) (x2,y2) are on the same line(Of course). 5 6 * Consider the gap between two points. 7 8 * We have (y2-y1)=a(x2-x1),a=(y2-y1)/(x2-x1) a is a rational, b is canceled since b is a constant 9 10 * If a third point (x3,y3) are on the same line. So we must have y3=ax3+b 11 12 * Thus,(y3-y1)/(x3-x1)=(y2-y1)/(x2-x1)=a 13 14 * Since a is a rational, there exists y0 and x0, y0/x0=(y3-y1)/(x3-x1)=(y2-y1)/(x2-x1)=a 15 16 * So we can use y0&x0 to track a line; 17 */ 18 19 public class Solution{ 20 public int maxPoints(Point[] points) { 21 if (points==null) return 0; 22 if (points.length<=2) return points.length; 23 24 Map<Integer,Map<Integer,Integer>> map = new HashMap<Integer,Map<Integer,Integer>>(); 25 int result=0; 26 for (int i=0;i<points.length;i++){ 27 map.clear(); 28 int overlap=0,max=0; 29 for (int j=i+1;j<points.length;j++){ 30 int x=points[j].x-points[i].x; 31 int y=points[j].y-points[i].y; 32 if (x==0&&y==0){ 33 overlap++; 34 continue; 35 } 36 int gcd=generateGCD(x,y); 37 if (gcd!=0){ 38 x/=gcd; 39 y/=gcd; 40 } 41 42 if (map.containsKey(x)){ 43 if (map.get(x).containsKey(y)){ 44 map.get(x).put(y, map.get(x).get(y)+1); 45 }else{ 46 map.get(x).put(y, 1); 47 } 48 }else{ 49 Map<Integer,Integer> m = new HashMap<Integer,Integer>(); 50 m.put(y, 1); 51 map.put(x, m); 52 } 53 max=Math.max(max, map.get(x).get(y)); 54 } 55 result=Math.max(result, max+overlap+1); 56 } 57 return result; 58 59 60 } 61 private int generateGCD(int a,int b){ 62 63 if (b==0) return a; 64 else return generateGCD(b,a%b); 65 66 } 67 }
参考：https://leetcode.com/problems/max-points-on-a-line/discuss/47113/A-java-solution-with-notes
