[leetcode] 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.

https://oj.leetcode.com/problems/max-points-on-a-line/

思路：以每一个点位轴点，计算其他点的斜率，如果斜率相同则共线，用hashmap记录共线点数，不断更新最大值。

　　注意：有重复点需要处理，尤其是所有点都是重复点的情况需要特殊处理。

import java.util.HashMap;
import java.util.Map;

/**
* /article/1565212.html
* /article/4879662.html
* @author jd
*
*/
public class Solution {
public int maxPoints(Point[] points) {
if (points == null)
return 0;
int n = points.length;
if (n <= 2)
return n;
Map<Double, Integer> map = new HashMap<Double, Integer>();
int maxN = 0;

for (int i = 0; i < points.length; i++) {
map.clear();
Point cur = points[i];
int duplicates = 1;

boolean notAllDup = false;
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) {
duplicates++;
} else {
notAllDup = true;
Double slope = slope(cur, points[j]);
if (map.get(slope) == null) {
map.put(slope, 1);
} else {
map.put(slope, map.get(slope) + 1);
}
}
}

if (!notAllDup) {
maxN = Math.max(maxN, duplicates);
}

for (Double key : map.keySet()) {
maxN = Math.max(maxN, map.get(key) + duplicates);
}

}

return maxN;
}

private double slope(Point cur, Point p) {
int x1 = cur.x;
int y1 = cur.y;
int x2 = p.x;
int y2 = p.y;
if (x1 == x2 && y1 == y2) {
return -Double.MAX_VALUE;
} else if (x1 == x2)
return Double.MAX_VALUE;
else if (y1 == y2)
return 0;
else
return 1.0 * (y2 - y1) / (x2 - x1);

}

public static void main(String[] args) {
Point[] points = null;
// 3
points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(9, 10), new Point(10, 9) };
System.out.println(new Solution().maxPoints(points));

// 5
points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(10, 10), new Point(9, 9) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(0, 0), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 0) };
System.out.println(new Solution().maxPoints(points));

// 1
points = new Point[] { new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 2
points = new Point[] { new Point(1, 1), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 2
points = new Point[] { new Point(0, 0), new Point(1, 1), new Point(1, -1) };
System.out.println(new Solution().maxPoints(points));
}

}

第二遍记录：

　　对于求slope，注意斜率无限大的情况和两个点重合的情况。

　　统计点时，注意与当前点的重合点的统计，以及所有点都是重合点的特殊情况。

第三遍记录：

　　重新写了一边，对于每个节点依次统计跟其他节点的斜率数， 注意跟当前节点重合的点要分开统计。

import java.util.HashMap;

public class Solution {

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

int max = 2;
// at least 3 points now
for (int i = 0; i < points.length; i++) {
Point cur = points[i];
HashMap<Double, Integer> map = new HashMap<Double, Integer>();
int duplicates = 1;
int curMax = 0;
for (int j = 0; j < points.length; j++) {
if (i == j)
continue;
if (cur.x == points[j].x && cur.y == points[j].y) {
duplicates++;
} else {
double curSlope = slope(cur, points[j]);
if (map.containsKey(curSlope)) {
map.put(curSlope, map.get(curSlope) + 1);
} else {
map.put(curSlope, 1);
}
}

}

for (Double each : map.keySet()) {
curMax = Math.max(curMax, map.get(each));
}

max = Math.max(max, duplicates + curMax);

}

return max;
}

private double slope(Point a, Point b) {
int x1 = a.x;
int y1 = a.y;
int x2 = b.x;
int y2 = b.y;
if (x1 == x2 && y1 == y2) {
return Integer.MIN_VALUE;
} else if (y1 == y2) {
return 0;
} else if (x1 == x2) {
return Integer.MAX_VALUE;
} else {
return 1.0 * (y1 - y2) / (x1 - x2);
}

}

public static void main(String[] args) {
Point[] points = null;
// 3
points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(9, 10), new Point(10, 9) };
System.out.println(new Solution().maxPoints(points));

// 5
points = new Point[] { new Point(1, 1), new Point(2, 2), new Point(3, 3), new Point(10, 10), new Point(9, 9) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(0, 0), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 3
points = new Point[] { new Point(1, 1), new Point(1, 1), new Point(1, 0) };
System.out.println(new Solution().maxPoints(points));

// 1
points = new Point[] { new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 2
points = new Point[] { new Point(1, 1), new Point(1, 1) };
System.out.println(new Solution().maxPoints(points));

// 2
points = new Point[] { new Point(0, 0), new Point(1, 1), new Point(1, -1) };
System.out.println(new Solution().maxPoints(points));
}

}

class Point {
int x;
int y;

public Point(int x, int y) {
this.x = x;
this.y = y;
}

}

参考：

/article/1565212.html

/article/4879662.html