[LeetCode][JavaScript]Max Points on a Line

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.

粗暴地用二重循环遍历。

每一轮都构造一个哈希表，用来记录斜率，斜率k = (y1 - y2) / (x1 - x2)。

注意到特殊情况：

1.两点重合，用countSamePoint记下重复的点，最后加到结果上。

2.两点与X轴平行，此时y1 - y2 = 0， k = 0，构造哈希表，提前塞一个slopeMap[0] = 0。

3.两点与Y轴平行，此时x1 - x2 = 0， k = Infinity，同上塞一个slopeMap[Infinity] = 0。

 1 /**
 2  * Definition for a point.
 3  * function Point(x, y) {
 4  *     this.x = x;
 5  *     this.x = y;
 6  * }
 7  */
 8 /**
 9  * @param {Point[]} points
10  * @return {number}
11  */
12 var maxPoints = function(points) {
13     if(points.length <= 1){
14         return points.length;
15     }
16     var res = -1, countSamePoint, max, i, j, slopeMap, curr, slope, tmp;
17     for(i = 0; i < points.length; i++){
18         curr = points[i];
19         max = 0;
20         countSamePoint = 1;
21         slopeMap = {};
22         slopeMap[0] = 0; slopeMap[Infinity] = 0;
23         for(j = 0; j < points.length; j++){
24             if(i === j){
25                 continue;
26             }
27             if(points[j].x === curr.x && points[j].y === curr.y){
28                 countSamePoint++;
29             }else{
30                 slope = (points[j].y - curr.y) / (points[j].x - curr.x);
31                 if(slopeMap[slope] === undefined){
32                     slopeMap[slope] = 1;
33                 }else{
34                     slopeMap[slope]++;
35                 }
36                 tmp = slopeMap[slope];
37                 if(tmp > max){
38                     max = tmp;
39                 }
40             }
41             if(max + countSamePoint > res){
42                 res = max + countSamePoint;
43             }
44         }
45     }
46     return res;
47 };

Test Cases:

 1 function test(){
 2     console.log(maxPoints([])); //0
 3     console.log(maxPoints([new Point(0,0)])); //1
 4     console.log(maxPoints([new Point(0,0),new Point(0,0)])); //2
 5     console.log(maxPoints([new Point(1,1),new Point(2,1)])); //2
 6     console.log(maxPoints([new Point(1,1),new Point(1,2)])); //2
 7     console.log(maxPoints([new Point(0,0),new Point(1,1),new Point(0,0)])); //3
 8     console.log(maxPoints([new Point(1,1),new Point(2,2),new Point(2,3),new Point(3,3),new Point(5,10)]));  //3
 9     console.log(maxPoints([new Point(1,1),new Point(1,1),new Point(2,2),new Point(2,2)])); //4
10
11 };