【LeetCode】149. Max Points on a Line - javascript

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;
};