三刷149. Max Points on a Line

Hard
关于samePoints, max, res之间的关系：

image.png

这里还要介绍一下如何计算两个整数的最大公约数，也就是gcd, 这里用的是欧几里得算法：

• 基本的Euclidean algorithms， 求a,b的gcd, 就等于求较小数和大数减小数的差的gcd, 知道有一个数等于0，剩下的另一个不为零的数就是gcd;
• 扩展的Euclidean algorithms，求a,b的gcd， 这次不减去小数，而是除以小数取余数，直到取到余数等于0时。

public int gcd(int a, int b){
  if (b == 0){
    return a;
  } else {
    return gcd(b, a%b);  
  }
}

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
class Solution {
    public int maxPoints(Point[] points) {
        if (points == null || points.length == 0){
            return 0;
        }
        Map map = new HashMap<>();
        int res = 0;
        for (int i = 0; i < points.length; i++){
            map = new HashMap<>();
            int samePoints = 1;
            int max = 0;
            for (int j = i + 1; j < points.length; j++){
                if (points[j].y == points[i].y && points[j].x == points[i].x){
                    samePoints++;
                    continue;
                }     
                int deltaY = points[j].y - points[i].y;
                int deltaX = points[j].x - points[i].x;
                int gcd = gcd(deltaY, deltaX);
                deltaY /= gcd;
                deltaX /= gcd;
                String slope = deltaY+"/"+deltaX;
                map.put(slope, map.getOrDefault(slope, 0) + 1);
                max = Math.max(max, map.get(slope));
            }
            res = Math.max(res, max + samePoints);
        }
        return res;
    }
        
    private int gcd(int a, int b){
        if (b == 0){
            return a;
        } else {
            return gcd(b, a % b);
        }
    }
}