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  • [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.


    求有多少点在一直线上。

    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 };
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  • 原文地址:https://www.cnblogs.com/Liok3187/p/4733404.html
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