Leetcode 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.

标签 Hash Table Math

类似题目 (M) Line Reflection

 

 

 

思路：注意可能存在重合点，除去重合点，只要记录能够表示一条直线的(x0,y0)，其中a=x1-x2和y1-y2的最大公约数，x0=(x1-x2)/a,y0=(y1-y2)/a。注意到对于特定的i0，只要第一次出现(x0,y0)，那么在直线(x0,y0)的点的数量一定可以在i0下得到最大值，本质上就是比较每个i下第一次出现的(x0,y0)包含的点的数目。

 

代码：

/**
 * 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) {
        HashMap<Integer, Map<Integer, Integer>> hashMap = new HashMap<Integer, Map<Integer, Integer>>();
        int res = 0, maxCount = 0, overlap = 0;
        for (int i = 0; i < points.length; ++i) {
            hashMap.clear();
            maxCount = overlap = 0;
            for (int j = i + 1; j < points.length; j++) {
                int x = points[i].x - points[j].x;
                int y = points[i].y - points[j].y;
                if (x == 0 && y == 0) {//重合点
                    overlap++;
                    continue;
                }
                int gcd = Gcd(x, y);
                x = x / gcd;
                y = y / gcd;
                if (hashMap.containsKey(x)) {
                    hashMap.get(x).put(y, hashMap.get(x).getOrDefault(y, 0) + 1);
                }else {
                    Map<Integer,Integer> map = new HashMap<Integer,Integer>();
                    map.put(y, 1);
                    hashMap.put(x, map);
                }
                maxCount = Math.max(maxCount, hashMap.get(x).get(y));
            }
            res = Math.max(res, maxCount + overlap + 1);
        }
        return res;
    }
    private int Gcd(int x, int y){
        if (y == 0) return x;
        return Gcd(y, x%y);
    }
}