149. Max Points on a Line

题目：

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

链接： http://leetcode.com/problems/max-points-on-a-line/

题解：

这道题是旧时代的残党，LeetCode大规模加新题之前的最后一题，新时代没有可以载你的船。要速度解决此题，之后继续刷新题。

主要思路是，对每一个点求其于其他点的斜率，放在一个HashMap中，每计算完一个点，尝试更新global max。

要注意的一些地方是 -

跳过当前点
处理重复
计算斜率的时候使用double
保存斜率时，第一次要保存2个点，并非1个
假如map.size() = 0，这时尝试用比较max和 duplicate + 1来更新max

Time Complexity - O(n²)， Space Complexity - O(n)

/**
 * 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) {
        if(points == null || points.length == 0)
            return 0;
        HashMap<Double, Integer> map = new HashMap<>();
        int max = 1;
        
        for(int i = 0; i < points.length; i++) {        //for each point, find out slopes to other points
            map.clear();
            int duplicate = 0;                          //dealing with duplicates
            
            for(int j = 0; j < points.length; j++) {
                if(j == i)
                    continue;
                if(points[i].x == points[j].x && points[i].y == points[j].y) {
                    duplicate++;    
                    continue;
                }
                    
                double slope = (double)(points[j].y - points[i].y) / (double)(points[j].x - points[i].x);
                
                if(map.containsKey(slope)) {
                    map.put(slope, map.get(slope) + 1);
                } else
                    map.put(slope, 2);
            }
            
            if(map.size() > 0) {
                for(int localMax : map.values())
                    max = Math.max(max, localMax + duplicate);    
            } else 
                max = Math.max(max, duplicate + 1);    
        }
        
        return max;
    }
}

二刷:

又做到了这一题。题目很短，很多东西没有说清楚。比如假如这n个点中有重复，这重复的点我们也要算在结果里。比如[0, 0][0, 0]这是两个点。

这回用的方法依然是双重循环，使用一个Map<Double, Integer>来记录每个点到其他点的斜率slope，然后每统计完一个点，我们尝试更新一下globalMax。这里有几点要注意:

有重复点的情况，这时候我们用一个整数dupPointNum来记录，并且跳过后面的运算
斜率
1. 当点p1.x == p2.x时这时我们要设置slope = 0.0，否则map里可能会出现-0.0和0.0两个key
2. 当点p1.y == p2.y的时候，我们要设置slope = Double.POSITIVE_INFINITY，否则map里可能出现Double.POSITIVE_INFINITY或者Double.NAGETIVE_INFINITY
3. 其他情况我们可以使用直线的两点式方程算出斜率slope = (double)(p1.y - p2.y) / (p1.x - p2.x)
计算完一个点之后，我们遍历Map的value()集，跟globalMax比较，尝试更新

有意思的一点是Double.NaN虽然不等于Double.NaN，即(Double.NaN == Double.NaN)返回false。但作为map的key来说却能在查找时返回true，所以2.2也可以设置slope = Double.NaN.

Java:

Time Complexity - O(n²)， Space Complexity - O(n)

/**
 * 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) {
        if (points == null || points.length == 0) return 0;
        Map<Double, Integer> map = new HashMap<>(points.length);
        int max = 0, len = points.length;
        
        for (int i = 0; i < len; i++) {
            map.clear();
            int dupPointNum = 0;
            for (int j = i + 1; j < len; j++) {
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    dupPointNum++;
                    continue;
                }
                double slope = 0.0;
                if (points[j].y == points[i].y) slope = 0.0;
                else if (points[j].x == points[i].x) slope = Double.POSITIVE_INFINITY;
                else slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);
                
                if (map.containsKey(slope)) map.put(slope, map.get(slope) + 1);
                else map.put(slope, 2);
            }
            max = Math.max(max, dupPointNum + 1);
            for (int count : map.values()) max = Math.max(max, count + dupPointNum);
        }
        return max;
    }
}

Update:

加入了排序去重，速度更快了一些。

/**
 * 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) {
        if (points == null || points.length == 0) return 0;
        Arrays.sort(points, (Point p1, Point p2)-> (p1.x != p2.x) ? p1.x - p2.x : p1.y - p2.y);
        Map<Double, Integer> map = new HashMap<>(points.length);
        int max = 0, len = points.length;
        
        for (int i = 0; i < len; i++) {
            if (i > 0 && points[i].x == points[i - 1].x && points[i].y == points[i - 1].y) continue;
            map.clear();
            int dupPointNum = 0;
            for (int j = i + 1; j < len; j++) {
                if (points[i].x == points[j].x && points[i].y == points[j].y) {
                    dupPointNum++;
                    continue;
                }
                double slope = 0.0;
                if (points[j].y == points[i].y) slope = 0.0;
                else if (points[j].x == points[i].x) slope = Double.POSITIVE_INFINITY;
                else slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);
                
                if (map.containsKey(slope)) map.put(slope, map.get(slope) + 1);
                else map.put(slope, 2);
            }
            max = Math.max(max, dupPointNum + 1);
            for (int count : map.values()) max = Math.max(max, count + dupPointNum);
        }
        return max;
    }
}

149. Max Points on a Line

Reference：

https://leetcode.com/discuss/72457/java-27ms-solution-without-gcd

https://leetcode.com/discuss/57464/accepted-java-solution-easy-to-understand

149. Max Points on a Line

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