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Leetcode149-maxPointsOnALine

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Example

Example 1:

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Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

Example 2:
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Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Solution

Solution 1: Hash Map

O(n^2), used (x1 - x2) / gcd and (y1 - y2) / gcd (gcd is the gcd between (x1 - x2) and (y1-y2))

To avoid to use double, which can occur accuracy problem

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class Solution {
    public int maxPoints(int[][] points) {
        if (points == null) return 0;
        if (points.length <= 2) return points.length;

        Map<Integer, Map<Integer, Integer>> map = new HashMap<Integer, Map<Integer, Integer>>();
        int result = 0;
        for (int i = 0; i < points.length; i++) {
            map.clear();
            int num = 1, max = 0;
            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];
                if (x == 0 && y == 0) {
                    num++;
                    continue;
                }
                int gcd = generateGCD(x, y);
                if (gcd != 0) {
                    x /= gcd;
                    y /= gcd;
                }

                if (map.containsKey(x)) {
                    map.get(x).put(y, map.get(x).getOrDefault(y, 0) + 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 + num);
        }
        return result;
    }

    private int generateGCD(int a, int b) {
        if (b == 0) return a;
        else return generateGCD(b, a % b);
    }
}

Solution 2

O(n^3), used Cross product to determine if these three points are in one line

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class Solution {
    public int maxPoints(int[][] points) {
        int res = 0, n = points.length;
        for (int i = 0; i < n; ++i) {
            int duplicate = 1;
            for (int j = i + 1; j < n; ++j) {
                int cnt = 0;
                long x1 = points[i][0], y1 = points[i][1];
                long x2 = points[j][0], y2 = points[j][1];
                if (x1 == x2 && y1 == y2) {
                    duplicate++;
                    continue;
                }
                for (int k = 0; k < n; ++k) {
                    int x3 = points[k][0], y3 = points[k][1];
                    if (x1*y2 + x2*y3 + x3*y1 - x3*y2 - x2*y1 - x1 * y3 == 0) {
                        cnt++;
                    }
                }
                res = Math.max(res, cnt);
            }
            res = Math.max(res, duplicate);
        }
        return res;
    }
}