9.8 KiB
id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5951a53863c8a34f02bf1bdc | 最近对的问题 | 1 | 302232 | closest-pair-problem |
--description--
Provide a function to find the closest two points among a set of given points in two dimensions.
直接的解决方案是 O(n^2) 的算法(我们可以称之为 brute-force 算法);伪代码(使用索引)可能很简单:
bruteForceClosestPair of P(1), P(2), ... P(N)
if N < 2 then
return ∞
else
minDistance ← |P(1) - P(2)|
minPoints ← { P(1), P(2) }
foreach i ∈ [1, N-1]
foreach j ∈ [i+1, N]
if |P(i) - P(j)| < minDistance then
minDistance ← |P(i) - P(j)|
minPoints ← { P(i), P(j) }
endif
endfor
endfor
return minDistance, minPoints
endif
更好的算法基于递归分治法,即 $O(n\log n)$,伪代码可以是:
closestPair of (xP, yP)
where xP is P(1) .. P(N) sorted by x coordinate, and
yP is P(1) .. P(N) sorted by y coordinate (ascending order)
if N ≤ 3 then
return closest points of xP using brute-force algorithm
else
xL ← points of xP from 1 to ⌈N/2⌉
xR ← points of xP from ⌈N/2⌉+1 to N
xm ← xP(⌈N/2⌉)x
yL ← { p ∈ yP : px ≤ xm }
yR ← { p ∈ yP : px > xm }
(dL, pairL) ← closestPair of (xL, yL)
(dR, pairR) ← closestPair of (xR, yR)
(dmin, pairMin) ← (dR, pairR)
if dL < dR then
(dmin, pairMin) ← (dL, pairL)
endif
yS ← { p ∈ yP : |xm - px| < dmin }
nS ← number of points in yS
(closest, closestPair) ← (dmin, pairMin)
for i from 1 to nS - 1
k ← i + 1
while k ≤ nS and yS(k)y - yS(i)y < dmin
if |yS(k) - yS(i)| < closest then
(closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
endif
k ← k + 1
endwhile
endfor
return closest, closestPair
endif
对于输入,期望参数是一个 Point 对象数组,其中 x 和 y 成员设置为数字。 返回一个包含 distance 和 pair (两个最近点的对)的键-值对的对象。
例如带有输入数组 points 的 getClosestPair:
const points = [
new Point(1, 2),
new Point(3, 3),
new Point(2, 2)
];
会返回:
{
distance: 1,
pair: [
{
x: 1,
y: 2
},
{
x: 2,
y: 2
}
]
}
注意: 按 x 值的递增顺序对 pair 数组进行排序。
--hints--
getClosestPair 应该是一个函数。
assert(typeof getClosestPair === 'function');
getClosestPair(points1).distance 应该是 0.0894096443343775。
assert.equal(getClosestPair(points1).distance, answer1.distance);
getClosestPair(points1).pair 应该是 [ { x: 7.46489, y: 4.6268 }, { x: 7.46911, y: 4.71611 } ]。
assert.deepEqual(
JSON.parse(JSON.stringify(getClosestPair(points1))).pair,
answer1.pair
);
getClosestPair(points2).distance 应该是 65.06919393998976。
assert.equal(getClosestPair(points2).distance, answer2.distance);
getClosestPair(points2).pair 应该是 [ { x: 37134, y: 1963 }, { x: 37181, y: 2008 } ]。
assert.deepEqual(
JSON.parse(JSON.stringify(getClosestPair(points2))).pair,
answer2.pair
);
getClosestPair(points3).distance 应该是 6754.625082119658。
assert.equal(getClosestPair(points3).distance, answer3.distance);
getClosestPair(points3).pair 应该是 [ { x: 46817, y: 64975 }, { x: 48953, y: 58567 } ]。
assert.deepEqual(
JSON.parse(JSON.stringify(getClosestPair(points3))).pair,
answer3.pair
);
--seed--
--after-user-code--
const points1 = [
new Point(0.748501, 4.09624),
new Point(3.00302, 5.26164),
new Point(3.61878, 9.52232),
new Point(7.46911, 4.71611),
new Point(5.7819, 2.69367),
new Point(2.34709, 8.74782),
new Point(2.87169, 5.97774),
new Point(6.33101, 0.463131),
new Point(7.46489, 4.6268),
new Point(1.45428, 0.087596)
];
const answer1 = {
distance: 0.0894096443343775,
pair: [
{
x: 7.46489,
y: 4.6268
},
{
x: 7.46911,
y: 4.71611
}
]
};
const points2 = [
new Point(37100, 13118),
new Point(37134, 1963),
new Point(37181, 2008),
new Point(37276, 21611),
new Point(37307, 9320)
];
const answer2 = {
distance: 65.06919393998976,
pair: [
{
x: 37134,
y: 1963
},
{
x: 37181,
y: 2008
}
]
};
const points3 = [
new Point(16910, 54699),
new Point(14773, 61107),
new Point(95547, 45344),
new Point(95951, 17573),
new Point(5824, 41072),
new Point(8769, 52562),
new Point(21182, 41881),
new Point(53226, 45749),
new Point(68180, 887),
new Point(29322, 44017),
new Point(46817, 64975),
new Point(10501, 483),
new Point(57094, 60703),
new Point(23318, 35472),
new Point(72452, 88070),
new Point(67775, 28659),
new Point(19450, 20518),
new Point(17314, 26927),
new Point(98088, 11164),
new Point(25050, 56835),
new Point(8364, 6892),
new Point(37868, 18382),
new Point(23723, 7701),
new Point(55767, 11569),
new Point(70721, 66707),
new Point(31863, 9837),
new Point(49358, 30795),
new Point(13041, 39744),
new Point(59635, 26523),
new Point(25859, 1292),
new Point(1551, 53890),
new Point(70316, 94479),
new Point(48549, 86338),
new Point(46413, 92747),
new Point(27186, 50426),
new Point(27591, 22655),
new Point(10905, 46153),
new Point(40408, 84202),
new Point(52821, 73520),
new Point(84865, 77388),
new Point(99819, 32527),
new Point(34404, 75657),
new Point(78457, 96615),
new Point(42140, 5564),
new Point(62175, 92342),
new Point(54958, 67112),
new Point(4092, 19709),
new Point(99415, 60298),
new Point(51090, 52158),
new Point(48953, 58567)
];
const answer3 = {
distance: 6754.625082119658,
pair: [
{
x: 46817,
y: 64975
},
{
x: 48953,
y: 58567
}
]
}
--seed-contents--
const Point = function(x, y) {
this.x = x;
this.y = y;
};
Point.prototype.getX = function() {
return this.x;
};
Point.prototype.getY = function() {
return this.y;
};
function getClosestPair(pointsArr) {
return true;
}
--solutions--
const Point = function(x, y) {
this.x = x;
this.y = y;
};
Point.prototype.getX = function() {
return this.x;
};
Point.prototype.getY = function() {
return this.y;
};
const mergeSort = function mergeSort(points, comp) {
if(points.length < 2) return points;
var n = points.length,
i = 0,
j = 0,
leftN = Math.floor(n / 2),
rightN = leftN;
var leftPart = mergeSort( points.slice(0, leftN), comp),
rightPart = mergeSort( points.slice(rightN), comp );
var sortedPart = [];
while((i < leftPart.length) && (j < rightPart.length)) {
if(comp(leftPart[i], rightPart[j]) < 0) {
sortedPart.push(leftPart[i]);
i += 1;
}
else {
sortedPart.push(rightPart[j]);
j += 1;
}
}
while(i < leftPart.length) {
sortedPart.push(leftPart[i]);
i += 1;
}
while(j < rightPart.length) {
sortedPart.push(rightPart[j]);
j += 1;
}
return sortedPart;
};
const closestPair = function _closestPair(Px, Py) {
if(Px.length < 2) return { distance: Infinity, pair: [ new Point(0, 0), new Point(0, 0) ] };
if(Px.length < 3) {
//find euclid distance
var d = Math.sqrt( Math.pow(Math.abs(Px[1].x - Px[0].x), 2) + Math.pow(Math.abs(Px[1].y - Px[0].y), 2) );
return {
distance: d,
pair: [ Px[0], Px[1] ]
};
}
var n = Px.length,
leftN = Math.floor(n / 2),
rightN = leftN;
var Xl = Px.slice(0, leftN),
Xr = Px.slice(rightN),
Xm = Xl[leftN - 1],
Yl = [],
Yr = [];
//separate Py
for(var i = 0; i < Py.length; i += 1) {
if(Py[i].x <= Xm.x)
Yl.push(Py[i]);
else
Yr.push(Py[i]);
}
var dLeft = _closestPair(Xl, Yl),
dRight = _closestPair(Xr, Yr);
var minDelta = dLeft.distance,
closestPair = dLeft.pair;
if(dLeft.distance > dRight.distance) {
minDelta = dRight.distance;
closestPair = dRight.pair;
}
//filter points around Xm within delta (minDelta)
var closeY = [];
for(i = 0; i < Py.length; i += 1) {
if(Math.abs(Py[i].x - Xm.x) < minDelta) closeY.push(Py[i]);
}
//find min within delta. 8 steps max
for(i = 0; i < closeY.length; i += 1) {
for(var j = i + 1; j < Math.min( (i + 8), closeY.length ); j += 1) {
var d = Math.sqrt( Math.pow(Math.abs(closeY[j].x - closeY[i].x), 2) + Math.pow(Math.abs(closeY[j].y - closeY[i].y), 2) );
if(d < minDelta) {
minDelta = d;
closestPair = [ closeY[i], closeY[j] ]
}
}
}
return {
distance: minDelta,
pair: closestPair.sort((pointA, pointB) => pointA.x - pointB.x)
};
};
function getClosestPair(points) {
const sortX = function(a, b) { return (a.x < b.x) ? -1 : ((a.x > b.x) ? 1 : 0); }
const sortY = function(a, b) { return (a.y < b.y) ? -1 : ((a.y > b.y) ? 1 : 0); }
const Px = mergeSort(points, sortX);
const Py = mergeSort(points, sortY);
return closestPair(Px, Py);
}