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freeCodeCamp/curriculum/challenges/chinese/22-rosetta-code/rosetta-code-challenges/closest-pair-problem.md
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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
  returnelse
  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 xy 成员设置为数字。 返回一个包含 distancepair (两个最近点的对)的键-值对的对象。

例如带有输入数组 pointsgetClosestPair

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: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);
}