[LeetCode]Maximum number of points on a straight line in 2d plane

最新推荐文章于 2019-07-23 23:42:12 发布

原创最新推荐文章于 2019-07-23 23:42:12 发布 · 522 阅读

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本文介绍了一种解决几何问题的方法：如何找出平面上给定点集中最大数量的共线点。通过计算每两点间的斜率并利用哈希映射记录出现次数最多的斜率来确定最长的共线点集合。

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Solution:

Remember that a line can be represented by y=kx+d, if p1 and p2 are in same line, then y1=x1k+d, y2=kx2+d, so y2-y1=kx2-kx1, so k=(y2-y1)/(x2-x1), then we can apply this formula to check if two points are in same line, however how to handle the duplicate points problem?

So instead to calculate the line with maximum points , we should calculate longest line(maximum ponts) through same point,

public class Solution {
     public  int maxPoints(Point[] points) {

            if (points==null||points.length==0){
                return 0;
            }  
            //slope -> number of points
            HashMap<Double, Integer> map=new HashMap<Double, Integer>();;
            int max=1;

            for(int i=0; i<points.length; i++){
                // current point changed, clear map for this point
                map.clear();

                // maybe all points contained in the list are same points
                // ,and same points' k is represented by Integer.MIN_VALUE
                map.put((double)Integer.MIN_VALUE, 1);

                int dup=0;

                for(int j=i+1; j<points.length; j++){
                   if (points[j].x==points[i].x&&points[j].y==points[i].y){
                       dup++;
                       continue;
                   }

                   // 1) look 0.0+(double)(points[j].y-points[i].y)/(double)(points[j].x-points[i].x)
                   // because (double)0/-1 is -0.0, so we should use 0.0+-0.0=0.0 to solve 0.0 !=-0.0 problem

                   // 2) if the line through two points are parallel to y coordinator, then K(slop) is Integer.MAX_VALUE
                   double key=points[j].x-points[i].x==0?Integer.MAX_VALUE:0.0+(double)(points[j].y-points[i].y)/(double)(points[j].x-points[i].x);

                   if (map.containsKey(key)){
                       map.put(key, map.get(key)+1);
                   }
                   else{
                      map.put(key, 2);
                   }
               }

              for (int temp: map.values()){
                  // duplicate may exist
                  //all duplicate of the cur point are on the same line
                  max = Math.max(max, temp+dup);
              }
            }
            return max;
        }
}