（leetcode)Max Points on a Line

Max Points on a Line

题目描述：

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

题目分析：

 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（谨记避免重复）

1 暴力求解：对于每两个点，求出与这两个点在同一直线的所有点，比较简单粗暴。

public class MaxPoints {/*public class Point {int x;int y;Point() { x = 0; y = 0; }Point(int a, int b) { x = a; y = b; }} */public static int maxPoints(Point[] points) {if(points==null||points.length<=0)return 0;if(points.length<=2)return points.length;int max=0;for(int i=0;i<points.length-1;i++){for(int j=i+1;j<points.length;j++){int count=2;//至少有两个在同一条直线上float k=0;if(points[i].x==points[j].x){k=Float.MAX_VALUE;}else{//求斜率公式int bigX=i;int smallX=j;if(points[i].x<points[j].x){bigX=j;smallX=i;}k=(float) (1.0*(points[bigX].y-points[smallX].y)/(points[bigX].x-points[smallX].x));}//这里从i而不是从就j的下一位开始，就是为了避免有重复项，比如（1,1）、（1,1）、（1,2）for(int next=i+1;next<points.length;next++){if(next==j)continue;float k2;if(points[next].x==points[j].x){k2=Float.MAX_VALUE;}else{int bigX=next;int smallX=j;if(points[next].x<points[j].x){bigX=j;smallX=next;}k2=(float) (1.0*(points[bigX].y-points[smallX].y)/(points[bigX].x-points[smallX].x));}if(k2==k)count++;if(k2!=k&&points[next].x==points[j].x&&points[next].y==points[j].y)count++;}if(count>max){max=count;//System.out.println(k);}}}return max;}

此算法复杂度为O(n^3),比较原始，另参考别人实现的算法，用空间换时间，使用HashMap记录斜率，实现了O(n^2)算法的

public int maxPoints(Point[] points) {          if(points.length == 0||points == null)             return 0;                      if(points.length <= 2)             return points.length;                      int max = 1;  //the final max value, at least one        for(int i = 0; i < points.length; i++) {              HashMap<Float, Integer> hm = new HashMap<Float, Integer>();              int same = 0;            int localmax = 1; //the max value of current slope, at least one            for(int j = 0; j < points.length; j++) {                  if(i == j)                     continue;                                      if(points[i].x == points[j].x && points[i].y == points[j].y){                    same++;                     continue;                }                                float slope = ((float)(points[i].y - points[j].y))/(points[i].x - points[j].x);                                 if(hm.containsKey(slope))                      hm.put(slope, hm.get(slope) + 1);                  else                      hm.put(slope, 2);  //two points form a line            }                        for (Integer value : hm.values())                   localmax = Math.max(localmax, value);                        localmax += same;              max = Math.max(max, localmax);          }          return max;     }