Leetcode-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.

A better solution:

  1 /**
  2  * Definition for a point.
  3  * class Point {
  4  *     int x;
  5  *     int y;
  6  *     Point() { x = 0; y = 0; }
  7  *     Point(int a, int b) { x = a; y = b; }
  8  * }
  9  */
 10 class Line{    
 11     //a/b = (y1-y2)/(x1-x2), c/d=(x1y2-x2y1)/(x1-x2)
 12     int a, b, c, d;
 13     String uid;
 14     Set<Point> pList;
 15     public Line(Point p1, Point p2){
 16         pList = new HashSet<Point>();
 17         a = p1.y-p2.y;
 18         b = p1.x-p2.x;
 19         c = p1.x*p2.y-p2.x*p1.y;
 20         d = p1.x - p2.x;
 21 
 22         if (a==0){
 23             a = 0;
 24             b = 1;
 25             c = p1.y;
 26             d = 1;            
 27         } else if (b==0){
 28             a = 1;
 29             b = 0;
 30             c = p1.x;
 31             d = 1;
 32         } else {
 33             int gcd = getGCD(a,b);
 34             a = a/gcd;
 35             b = b/gcd;
 36             if (b<0 && a>0){
 37                 a = -a;
 38                 b = -b;
 39             }
 40             gcd = getGCD(c,d);
 41             c = c/gcd;
 42             d = d/gcd;
 43             if (d<0 && c>0){
 44                 c = -c;
 45                 d = -d;
 46             }
 47         }
 48         //generate uid
 49         setUID();
 50     }
 51 
 52     public void setUID(){
 53         uid = a + "/" + b + "/" + c + "/" + d;
 54     }
 55 
 56     public String getUID(){
 57         return uid;
 58     }
 59 
 60     private int getGCD(int x, int y){
 61         while (y!=0){
 62             int val = x % y;
 63             x = y;
 64             y = val;
 65         } 
 66         return x;
 67     }
 68 }
 69 
 70 public class Solution {
 71     public int maxPoints(Point[] points) {
 72         if (points.length<3) return points.length;
 73 
 74         int res = 0;
 75         Map<Point,Integer> pMap = new HashMap<Point,Integer>();
 76         for (int i=0;i<points.length;i++)
 77             if (pMap.containsKey(points[i])){
 78                 pMap.put(points[i],pMap.get(points[i])+1);
 79             } else {
 80                 pMap.put(points[i],1);
 81             }
 82 
 83         Map<String,Line> lineMap = new HashMap<String,Line>();
 84         Iterator iter1 = pMap.entrySet().iterator();
 85         while (iter1.hasNext()){
 86             Map.Entry en1 = (Map.Entry) iter1.next();
 87             int val = (int) en1.getValue();
 88             if (res<val) res = val;
 89             Point p1 = (Point) en1.getKey();
 90             Map<Point,Integer> pMap2 = new HashMap<Point,Integer>();            
 91             pMap2.putAll(pMap);
 92             pMap2.remove(p1);            
 93             while (!pMap2.isEmpty()){
 94                 Iterator iter2 = pMap2.entrySet().iterator();
 95                 Map.Entry en2 = (Map.Entry) iter2.next();
 96                 Point p2 = (Point) en2.getKey();
 97                 Line l1 = new Line(p1,p2);
 98                 String uid = l1.getUID();
 99                 if (lineMap.containsKey(uid)){
100                     Line l2 = lineMap.get(uid);
101                     l2.pList.add(p1);
102                     l2.pList.add(p2);
103                     for (Point p: l2.pList) 
104                         pMap2.remove(p);
105                 } else {
106                     l1.pList.add(p1);
107                     l1.pList.add(p2);
108                     pMap2.remove(p2);
109                     lineMap.put(l1.getUID(),l1);
110                 }
111             }
112         }
113 
114         iter1 = lineMap.entrySet().iterator();
115         while (iter1.hasNext()){
116             Map.Entry en = (Map.Entry) iter1.next();
117             Line l1 = (Line) en.getValue();
118             int pNum = 0;
119             for (Point p: l1.pList)
120                 pNum += pMap.get(p);
121             if (res<pNum) res = pNum;
122         }
123         return res;
124     }
125 }

Solution:

  1 /**
  2  * Definition for a point.
  3  * class Point {
  4  *     int x;
  5  *     int y;
  6  *     Point() { x = 0; y = 0; }
  7  *     Point(int a, int b) { x = a; y = b; }
  8  * }
  9  */
 10 
 11 class Line {
 12     int slopeX;
 13     int slopeY;
 14     int b;
 15     boolean negSlope;
 16     boolean samePointLine;
 17     Set<Point> pointSet;
 18     String uid;
 19     public Line(int sx, int sy, int bb,boolean spl){
 20         if (sx<0) slopeX = -sx;
 21         else slopeX = sx;
 22 
 23         if (sy<0) slopeY = -sy;
 24         else slopeY = sy;
 25 
 26         if (sy==0) slopeX=1;
 27         if (sx==0) slopeY=1;
 28         b = bb;
 29 
 30         if (sx<0 || sy<0) negSlope = true;
 31         else negSlope = false;
 32         
 33         samePointLine = spl;
 34 
 35         pointSet = new HashSet<Point>();
 36 
 37         uid = Integer.toString(slopeX)+"/"+Integer.toString(slopeY)+"/"+Double.toString(b)+"/"+Boolean.toString(negSlope)+"/"+Boolean.toString(samePointLine);
 38   }
 39 
 40   public boolean equals(Line l2){
 41       if (uid==l2.uid)
 42           return true;
 43       else return false;
 44   }
 45 
 46 }       
 47 
 48 
 49 public class Solution {
 50     public int maxPoints(Point[] points) {
 51         if (points.length==1) return 1;
 52         
 53         Set<Point> pSet = new HashSet<Point>();
 54         for (int i=0;i<points.length;i++) pSet.add(points[i]);
 55 
 56         Map<String,Line> lineList = new HashMap<String,Line>();
 57         List<String> uidList = new ArrayList<String>();
 58         for (int i=1;i<points.length;i++){
 59             Set<Point> curPSet = new HashSet<Point>();
 60             curPSet.addAll(pSet);
 61             for (int j=0;j<i;j++)
 62                 if (curPSet.contains(points[j])){
 63                 //This point has not been checked.
 64                 //Calculate the line formed by i and j
 65                     int ix = points[i].x;
 66                     int iy = points[i].y;
 67                     int jx = points[j].x;
 68                     int jy = points[j].y;
 69                     int sx = points[i].x-points[j].x;
 70                     int sy = points[i].y-points[j].y;
 71                     Line l1 = null;
 72                     //infnite slope line.
 73                     if (sx==0 && sy==0)
 74                         l1 = new Line(points[i].x,points[i].y,0,true);
 75                     else if (sx==0)
 76                         l1 = new Line(0,1,points[i].x,false);
 77                     else if (sy==0)
 78                         l1 = new Line(1,0,points[i].y,false);
 79                     else {
 80                         int bb = jy*ix-jx*iy;
 81                         int gcd = getGCD(sx,sy);
 82                         int gcd2 = getGCD(bb,sx);
 83                         sx = sx/gcd;
 84                         sy = sy/gcd;
 85                         bb = bb/gcd2;
 86                         l1 = new Line(sx,sy,bb,false);
 87                     }
 88 
 89                     //Check each exsiting line.
 90                     if (lineList.containsKey(l1.uid)){
 91                         lineList.get(l1.uid).pointSet.add(points[i]);
 92                         lineList.get(l1.uid).pointSet.add(points[j]);
 93                         curPSet.removeAll(lineList.get(l1.uid).pointSet);
 94                         
 95                     } else {                        
 96                         lineList.put(l1.uid,l1);
 97                         uidList.add(l1.uid);
 98                         l1.pointSet.add(points[i]);
 99                         l1.pointSet.add(points[j]);
100                     }
101                }
102         }
103 
104         int max = 0;
105         for (int i=0;i<uidList.size();i++)
106             if (lineList.get(uidList.get(i)).pointSet.size()>max)
107                 max = lineList.get(uidList.get(i)).pointSet.size();
108 
109         return max;
110         
111     }
112 
113 
114     public int getGCD(int a, int b){
115         int left = a%b;
116         while (left!=0){
117             a=b;
118             b=left;
119             left=a%b;
120         }
121         return b;
122     }
123 
124 
125 }
原文地址:https://www.cnblogs.com/lishiblog/p/4166259.html