【LeetCode】149. Max Points on a Line

提交效率：192 ms ，55%

这一题，出现了 javascript 精度问题。最后的案例 [[0,0],[94911151,94911150],[94911152,94911151]] 是没法通过的。会将会两个点计算为同一个斜率的，造成结果的不对。

94911150 / 94911151 = 0.9999999894638303 == 94911151 / 94911152

var maxPoints = function(points) {    if(points.length<2 ) return points.length;    var result = {"ins":0};    var max = 0;    for(var j = 0 ; j < points.length; j++){        result = {"ins":0},base = 1,onthis = false;        for(var i = 0, temp; i< points.length; i++){            if(points[i].x == points[j].x){                // 每多一个同点，所有点数加1                if(points[i].y == points[j].y){                    for(var key in result){                        result[key]++;                    }                    max = onthis? max+1:max;                    base++;                }else{                    result["ins"]++;                }               max = result["ins"] > max ? result["ins"] : max;                continue;            }            temp = (points[i].y - points[j].y)/(points[i].x - points[j].x);            result[temp] ? (result[temp]++):(result[temp] = base);                result[temp] > max ? ( max = result[temp] ,onthis = true) : null;        }     }     return max;};

参考排名第一的方法，使用其它方式表示斜率：

var maxPoints = function(points) {    let size = points.length;    if (size < 2) return size;    const gcd = (a, b) => {        if (b === 0) return a;        return gcd(b, a%b);    }    let res = 0;    for (let i = 0; i < size; ++i) {        let hash = new Map();        let horizontal = 0, vertical = 0, localMax = 0, overlap = 1;        for (let j = i + 1; j < size; ++j) {            let a = points[i], b = points[j];            if (a.x === b.x && a.y === b.y) ++overlap;            else if (a.x === b.x) ++vertical;            else if (a.y === b.y) ++horizontal;            else {                let dx = a.x - b.x;                let dy = a.y - b.y;                let d = gcd(dx, dy);                dx = Math.floor(dx / d);                dy = Math.floor(dy / d);                let pair = "" + dx + "," + dy;                let curPoints = hash.has(pair) ? hash.get(pair) + 1 : 1;                hash.set(pair, curPoints);                localMax = Math.max(localMax, curPoints);            }        }        res = Math.max(res, overlap + Math.max(localMax, Math.max(vertical, horizontal)));    }    return res;};