LeetCode--149.Max Points on a Line(一条线上最多有几个点)

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

 

Point数据结构

class Point {
    int x;
    int y;
    Point() {
        x = 0; y = 0;
    }
    Point(int a, int b) {
        x = a; y = b;
    }
}

方法：选定一个点分别和其他点求斜率，如果斜率相同则在一条直线上，记录下数量，不同的斜率的数量使用HashMap进行保存。然后再选择另一个点进行相同的操作。

注意：斜率为0，点重合的情况

public static int maxPoints(Point[] points) {
    int len = points.length;
    if(len<=2) return len;
    int maxnum = 0;
    HashMap<Double,Integer>  map = new HashMap<Double,Integer>();        
    for(int i = 0 ; i < len ; i++){
        map = new HashMap<Double ,Integer>();
        int maxnum_curr = 0;
        int samenum = 1;
        Point currp = points[i];            
        for(int j = i+1 ; j < len ; j++){
            double k = 0 ;
            if(points[j].x ==currp.x){ 
                if(points[j].y==currp.y){ //两个点是同一个点
                    samenum++;
                    continue;
                }else{//x坐标相等，斜率不存在
                    k = (double) Integer.MAX_VALUE;
                }
            }else{                    
                k = (points[j].y-currp.y)/(double)(points[j].x-currp.x);
                if(k==-0.0 ) k = 0;                     
            }
            
            if(!map.containsKey(k)){
                map.put(k,1);    
                if(maxnum_curr<1)
                    maxnum_curr = 1;
            }else{
                int knum = map.get(k)+1;
                map.put(k,knum);
                if(knum>maxnum_curr)
                    maxnum_curr = knum;
            }
        }            
        
        maxnum_curr = maxnum_curr+samenum;
        if( maxnum_curr>maxnum)
            maxnum = maxnum_curr;
    }
    return maxnum;
}