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.

分析：

题目是给定一个二维平面上的点，希望求得这个平面上经过最多点的直线经过了多少个点。解这个题目需要用到穷举的思想，即计算每个点到其他点的斜

率，如果斜率相同，则共线。按照这个思路我们可以创建一个键为double，值为int类型的HashMap，穷举每两个点之间的斜率时，判断HashMap中是否

存在该斜率，没有就创建这个斜率的键值对，有就将值加1。需要考虑与该点重合还有垂直的情况。

具体代码如下：

/**
* Definition for a point.
* class Point {
* int x;
* int y;
* Point() { x = 0; y = 0; }
* Point(int a, int b) { x = a; y = b; }
* }
*/

import java.util.HashMap;
public class Solution {
public int maxPoints(Point[] points) {
int len = points.length;

if (len < 2)
return len;

int ret = 0;
for (int i = 0; i < len; i++){
int chuizhi = 0;
int chonghe = 1;
HashMap<Double, Integer> hm = new HashMap<Double, Integer>();
for (int j = 0; j < len; j++){
if (i == j)
continue;
else{
int diffX = points[i].x - points[j].x;
int diffY = points[i].y - points[j].y;

if (diffX == 0 && diffY == 0){
chonghe++;
}
else if (diffX == 0){
chuizhi++;
}
else{
double slope = diffY*1.0/diffX;
if (hm.get(slope) == null){
hm.put(slope, 1);
}
else{
hm.put(slope, hm.get(slope)+1);
}

}
}

}

int max = chuizhi;
for(double k : hm.keySet()){
max = Math.max(max, hm.get(k));
}

ret = Math.max(ret, max+chonghe);
}

return ret;
}

}