freeCodeCamp/curriculum/challenges/chinese/22-rosetta-code/rosetta-code-challenges/closest-pair-problem.md

---
id: 5951a53863c8a34f02bf1bdc
title: 最近对的问题
challengeType: 1
forumTopicId: 302232
dashedName: 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：

<pre><strong>bruteForceClosestPair</strong> of P(1), P(2), ... P(N)
<strong>if</strong> N &#x3C; 2 <strong>then</strong>
  <strong>return</strong> ∞
<strong>else</strong>
  minDistance ← |P(1) - P(2)|
  minPoints ← { P(1), P(2) }
  <strong>foreach</strong> i ∈ [1, N-1]
    <strong>foreach</strong> j ∈ [i+1, N]
      <strong>if</strong> |P(i) - P(j)| &#x3C; minDistance <strong>then</strong>
        minDistance ← |P(i) - P(j)|
        minPoints ← { P(i), P(j) }
      <strong>endif</strong>
    <strong>endfor</strong>
  <strong>endfor</strong>
  <strong>return</strong> minDistance, minPoints
<strong>endif</strong>
</pre>

更好的算法基于递归分治法，即 $O(n\log n)$，伪代码可以是：

<pre><strong>closestPair</strong> 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)
<strong>if</strong> N ≤ 3 <strong>then</strong>
  <strong>return</strong> closest points of xP using brute-force algorithm
<strong>else</strong>
  xL ← points of xP from 1 to ⌈N/2⌉
  xR ← points of xP from ⌈N/2⌉+1 to N
  xm ← xP(⌈N/2⌉)<sub>x</sub>
  yL ← { p ∈ yP : p<sub>x</sub> ≤ xm }
  yR ← { p ∈ yP : p<sub>x</sub> > xm }
  (dL, pairL) ← closestPair of (xL, yL)
  (dR, pairR) ← closestPair of (xR, yR)
  (dmin, pairMin) ← (dR, pairR)
  <strong>if</strong> dL &#x3C; dR <strong>then</strong>
    (dmin, pairMin) ← (dL, pairL)
  <strong>endif</strong>
  yS ← { p ∈ yP : |xm - p<sub>x</sub>| &#x3C; dmin }
  nS ← number of points in yS
  (closest, closestPair) ← (dmin, pairMin)
  <strong>for</strong> i <strong>from</strong> 1 <strong>to</strong> nS - 1
    k ← i + 1
    <strong>while</strong> k ≤ nS <strong>and</strong> yS(k)<sub>y</sub> - yS(i)<sub>y</sub> &#x3C; dmin
      <strong>if</strong> |yS(k) - yS(i)| &#x3C; closest <strong>then</strong>
        (closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
      <strong>endif</strong>
      k ← k + 1
    <strong>endwhile</strong>
  <strong>endfor</strong>
  <strong>return</strong> closest, closestPair
<strong>endif</strong>
</pre>

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`：

```js
const points = [
  new Point(1, 2),
  new Point(3, 3),
  new Point(2, 2)
];
```

会返回：

```js
{
  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。

```js
assert(typeof getClosestPair === 'function');
```

`getClosestPair(points1).distance` 应该是 `0.0894096443343775`。

```js
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 } ]`。

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points1))).pair,
  answer1.pair
);
```

`getClosestPair(points2).distance` 应该是 `65.06919393998976`。

```js
assert.equal(getClosestPair(points2).distance, answer2.distance);
```

`getClosestPair(points2).pair` 应该是 `[ { x: 37134, y: 1963 }, { x: 37181, y: 2008 } ]`。

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points2))).pair,
  answer2.pair
);
```

`getClosestPair(points3).distance` 应该是 `6754.625082119658`。

```js
assert.equal(getClosestPair(points3).distance, answer3.distance);
```

`getClosestPair(points3).pair` 应该是 `[ { x: 46817, y: 64975 }, { x: 48953, y: 58567 } ]`。

```js
assert.deepEqual(
  JSON.parse(JSON.stringify(getClosestPair(points3))).pair,
  answer3.pair
);
```

# --seed--

## --after-user-code--

```js
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--

```js
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--

```js
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);
}
```