题解 | #直线上的牛#

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param points int整型二维数组
     * @return int整型
     */
    public int maxPoints (int[][] points) {
        // write code here
        if (points.length <= 2) {
            return points.length;
        }
        int result = 2;
        for (int i = 0; i < points.length-2; i++) {
            for (int j = i + 1; j < points.length-1; j++) {
                int count = 2;
                for (int k = j + 1; k < points.length; k++) {
                    if (points[i][0] == points[j][0]) {
                        if (points[k][0] == points[j][0]) {
                            count++;
                        }
                    } else if (points[i][1] == points[j][1]) {
                        if (points[k][1] == points[j][1]) {
                            count++;
                        }
                    } else {
                        if ((((points[k][1] - points[j][1]) / (points[k][0] - points[j][0])) == ((
                                    points[i][1] - points[j][1]) / (points[i][0] - points[j][0])))) {
                            count++;
                        }
                    }
                }
                result = Math.max(result, count);
            }
        }
        return result;
    }
}

本题我采用的是暴力匹配，所用编程语言是java。

先随便找两个点构成一条直线，再看剩下的点有多少个可以与之前的两个点构成同一条直线