g0101_0200.s0149_max_points_on_a_line.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0149_max_points_on_a_line;

// #Hard #Top_Interview_Questions #Array #Hash_Table #Math #Geometry #Algorithm_II_Day_21_Others
// #2022_06_24_Time_11_ms_(99.21%)_Space_41.5_MB_(95.53%)

/**
 * 149 - Max Points on a Line\.
 *
 * Hard
 *
 * Given an array of `points` where points[i] = [x_i, y_i] represents a point on the **X-Y** plane, return _the maximum number of points that lie on the same straight line_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/02/25/plane1.jpg)
 *
 * **Input:** points = \[\[1,1],[2,2],[3,3]]
 *
 * **Output:** 3 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/02/25/plane2.jpg)
 *
 * **Input:** points = \[\[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
 *
 * **Output:** 4 
 *
 * **Constraints:**
 *
 * *   `1 <= points.length <= 300`
 * *   `points[i].length == 2`
 * *   -10⁴ <= x_i, y_i <= 10⁴
 * *   All the `points` are **unique**.
**/
public class Solution {
    public int maxPoints(int[][] points) {
        if (points.length < 2) {
            return points.length;
        }
        int max = 0;
        for (int i = 0; i < points.length; i++) {
            for (int j = i + 1; j < points.length; j++) {
                int x = points[j][0] - points[i][0];
                int y = points[j][1] - points[i][1];
                int c = x * points[j][1] - y * points[j][0];
                int count = 2;
                for (int k = j + 1; k < points.length; k++) {
                    if (c == x * points[k][1] - y * points[k][0]) {
                        count++;
                    }
                }
                max = Math.max(count, max);
            }
        }
        return max;
    }
}