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[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.


Adjusted Difficulty 5
Time to use --------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)


This is a difficult coding question.

The idea is simple. For n points, there are n * (n-1) lines. Check slopes and then count total, we would get the answer.

However, coding of this idea is very difficult.


Firstly, there are 2 special cases when calculating the slope. The 2 points may locate in same position. And when point1.x = point2.x, slope = infinity. It’s easy to omit these 2 cases and result in mistake.

Secondly, when we count, we declare 2 variables: samePointNumber and maxPointCountWithSameSlope. It’s very important to initialize both values to 1 instead of 0! Why? Because these values just can’t be 0. I failed my 2nd version code when input = {(0,0), (0,0)}, the program shows result of 0, instead of 2.

Thirdly, about what data structure to use for counting. There is a discussion about this at here

  1. storing the vertical slopes as Double.NaN. That allows Double to represent every slope uniquely as (y/x).
  2. It is unsafe using floating points to make a hash, and -0.0 != 0.0

It’s great that using Double.NaN, it saves us time and effort to count vertical points. Second point is very valid, but it turns out that using HashMap<Double, Integer> can AC.

P.S. It is always not a good practise to use Double as hash key. See here.

Fourthly, I made a mistake here:

double slope = (p.y - q.y) / (p.x - q.x);

And it’s wrong. Why? Note that Point.x and Point.y are both integers. Integer division will return integer. We must cast it.

double slope = (double) (p.y - q.y) / (p.x - q.x);

Last, OMG I wish this is last, but not least, we can reduce execution time to half by checking only the points with larger index than the anchor point (that’s the name for ‘current point’). Good idea, right?

One more thing, how to iterate thru the HashMap (value only)? There is an easy way:

for (Integer a : map.values()) {

That’s all I’ve found for now.

Updated on Aug 12th, 2014

Based on the solution given in CC150 v4 Q10.6 on Page 199, it’s a proper way to solve with HashMap<Line, Integer> instead of using HashMap<Double, Integer>.

The reason is mentioned, it’s ‘unsafe using floating points to make a hash’.

Note that if we were to write our own ‘Line’ class, we must override the 2 methods:

  1. public int hashCode() {}
  2. public boolean equals(Object o) {}


written by me, Version3 using HashMap

public int maxPoints(Point[] points) {
    if (points.length <= 1)
        return points.length;
    HashMap<Double, Integer> map = null;
    int totalMax = 0;
    for (Point p : points) {
        int samePoint = 1;
        map = new HashMap<Double, Integer>();
        for (Point q : points) {
            if (q == p || p.x > q.x) {
            } else if (p.x == q.x && p.y == q.y) {
            } else {
                double slope = Double.NaN;
                if (p.x != q.x) {
                    slope = (double) (p.y - q.y) / (p.x - q.x);
                if (!map.containsKey(slope)) {
                    map.put(slope, 1);
                map.put(slope, map.get(slope) + 1);
        int pointMax = 1;
        for (Integer a : map.values()) {
            pointMax = Math.max(pointMax, a);
        totalMax = Math.max(totalMax, pointMax + samePoint - 1);
    return totalMax;