149. 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.

本题是比较难的题目，首先需要弄懂一点，就是最大公约数是最大的可整除数，也就是默认为k值（y=kx+b），然后用两个hashmap来存储（x，map），（y，number）值，再这里面要排除掉overlap的值，因此还需要用一个overlap的变量来记录相同的值。当然，本身的点也没有算进去，因此，总的result=max+1+overlap。代码如下：

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    public int maxPoints(Point[] points) {
        Map<Integer,Map<Integer,Integer>> map = new HashMap<>();
        if(points==null) return 0;
        int result = 0;
        if(points.length<=2) return points.length;
        for(int i=0;i<points.length;i++){
            map.clear();
            int max  =0;
            int overlap = 0;
            for(int j=i+1;j<points.length;j++){
                int x = points[j].x-points[i].x;
                int y = points[j].y-points[i].y;
                if(x==0&&y==0){
                    overlap++;
                    continue;
                } 
                int gcd = GCD(x,y);
                if(gcd!=0){
                    x = x/gcd;
                    y = y/gcd;   
                }
                if(map.containsKey(x)){
                    if(map.get(x).containsKey(y)){
                        map.get(x).put(y,map.get(x).get(y)+1);
                    }else{
                        map.get(x).put(y,1);
                    }
                }else{
                    Map<Integer,Integer> m = new HashMap<Integer,Integer>();
                    m.put(y,1);
                    map.put(x,m);
                }
                max = Math.max(max,map.get(x).get(y));
            }
            result = Math.max(result,max+1+overlap);
        }
        return result;
    }
    public int GCD(int a,int b){
        if(b==0) return a;
        return GCD(b,a%b);
    }
}

做这道题目要注意gcd不能为0；还有x=0，y=0的时候就要跳出循环；