Leetcode: 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.

解决：

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import java.util.HashMap;

import java.util.Map;

class Point {

int x;

int y;

Point() {

x = 0;

y = 0;

}

Point(int a, int b) {

x = a;

y = b;

}

}

/**

* 暴力破解：

* 遍历所有的点，每次取一个点，计算这个点与其他点斜率k，对k统计，找最大的。

* 注意相同的点， 水平或竖直同线的点。

* @author Administrator

*

*/

public class LinePoints {

public int maxPoints(Point[] points) {

if (points.length < 2)

return points.length;

int global_max = 0;

Map<Double, Integer> map_k = new HashMap<Double, Integer>();

for (int i = 0; i < points.length; i++) {//遍历，每次取一点A

int one_point_max = 0;

int h_max = 0;

int same_point_num = 0;

map_k.clear();

for (int j = 0; j < points.length; j++) {//计算点A与其他点的斜率k

if (i == j) {// itself

continue;

} else {

if (points[i].x == points[j].x

&& points[i].y == points[j].y) {// same point

same_point_num++;

continue;

} else {

if (points[i].x == points[j].x) {// 竖直同线的点

h_max++;

if (h_max > one_point_max)

one_point_max = h_max;

} else {

double k = 0;

if (points[i].y == points[j].y)//水平同线的点

k = 0;

else

k = ((double) (points[i].y - points[j].y))

/ ((double) (points[i].x - points[j].x));

int k_v = 1;

Double dk = new Double(k);

if (map_k.get(dk) != null) {

k_v = map_k.get(dk) + 1;

}

map_k.put(dk, new Integer(k_v));

if (k_v > one_point_max)

one_point_max = k_v;

}

}

}

}// end of for

if (one_point_max + same_point_num > global_max)

global_max = one_point_max + same_point_num;

}// end of for

return global_max + 1;

}

public static void main(String[] args) {

LinePoints lp = new LinePoints();

Point[] points = new Point[2];

points[0] = new Point(0, 0);

points[1] = new Point(0, 0);

int s = lp.maxPoints(points);

System.out.println(s);

}

}

参考：

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