题目：Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 题目是寻找平面中所有点在同一条直线上的最大个数。一种简单的思路就是使用Map来保存每条直线上的点数，然后遍历数组中的每个点即可。代码如下所示：
import java.util.HashMap; import java.util.Map; public class maxPoints { static class Point { int x; int y; Point() { x = 0; y = 0; } Point(int a, int b) { x = a; y = b; } } public int maxPoints(Point[] points) { if(points == null || points.length == 0) return 0; int max = 0; for(int i=0;i<points.length-1;i++){ //对各个点都初始化一个Map来保存该点所确定的所有直线上的点数 HashMap<Double, Integer> pointsOnLine = new HashMap<>(); //这两个变量分别保存相同点个数和x相同点的个数 int samePoint = 0, localMax = 0; //对每个点而言，遍历所有其他点 for(int j=i+1;j<points.length;j++){ int x = points[j].x - points[i].x; int y = points[j].y - points[i].y; if(x == 0 && y== 0) { samePoint++; continue; } //计算两个点确定的直线的斜率。只要斜率相同就说明该店在该直线上 double ratio = findRatio(x,y); if(!pointsOnLine.containsKey(ratio)) pointsOnLine.put(ratio, 1); //如果该斜率已经存在于map中，那么将其计数加一 else pointsOnLine.put(ratio, pointsOnLine.get(ratio)+1); localMax = Math.max(localMax, pointsOnLine.get(ratio)); } max = Math.max(max, localMax + samePoint); } return max+1; } private Double findRatio(int x, int y){ if(y == 0) return Double.MAX_VALUE; else return (double)x/(double) y + 0.0; } public static void main(String[] args){ Point p0 = new Point(0, 0); Point p1 = new Point(1, 1); Point p2 = new Point(1, -1); Point [] points = {p0, p1, p2}; } }