给定二维平面上的n个点,找出位于同一直线上的点的最大数目。
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
public class Solution {
public int maxPoints(Point[] points) {
//验证一下输入
if (points == null || points.length == 0){
return 0;
}
//一个点或者两个点直接返回值
if (points.length < 3){
return points.length;
}
int max = 2;
int sameSlopePoint;//相同斜率的点
int samePoint;//相同的点
for (int i = 0; i < points.length; i++){
//初始化一下,刚开始相同点都为零
samePoint = 0;
for(int j = i+1; j < points.length; j++){
sameSlopePoint = 1;//两点确定一条直线,这个点也应该纳入总数
int XDistance = points[j].x - points[i].x;
int YDistance = points[j].y - points[i].y;
if (XDistance == 0 && YDistance ==0){//特殊情况:与第一个点同一位置的点,这时候不用计算后面,因为两点在同一位置,不能确定一条直线
samePoint++;
}else{
sameSlopePoint = 2;
for(int k = j+1; k < points.length; k++){//找剩下斜率相同的点,三点的斜率相同则肯定在一条直线上
int XDistance2 = points[k].x - points[i].x;
int YDistance2 = points[k].y - points[i].y;
if(XDistance * YDistance2 == XDistance2 * YDistance){//相乘可以避免斜率为零导致除法异常
sameSlopePoint++;
}
}
}
if (max < sameSlopePoint + samePoint){
max = sameSlopePoint + samePoint;
}
}
}
return max;
}
}