chore(i8n,learn): processed translations
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Mrugesh Mohapatra
Mrugesh Mohapatra
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---
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id: 5951a53863c8a34f02bf1bdc
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title: 最近对的问题
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title: Closest-pair problem
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challengeType: 5
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videoUrl: ''
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forumTopicId: 302232
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dashedName: closest-pair-problem
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---
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# --description--
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任务：
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Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the [Closest pair of points problem](https://en.wikipedia.org/wiki/Closest pair of points problem "wp: Closest pair of points problem") in the *planar* case.
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提供一个函数来在二维中找到一组给定点中最接近的两个点，即求解平面情况下的[最近点对问题](<https://en.wikipedia.org/wiki/Closest pair of points problem> "wp：最近点的问题") 。
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The straightforward solution is a O(n<sup>2</sup>) algorithm (which we can call *brute-force algorithm*); the pseudo-code (using indexes) could be simply:
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直接的解决方案是O（n <sup>2</sup> ）算法（我们可以称之为强力算法）;伪代码（使用索引）可以简单地：
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<pre><strong>bruteForceClosestPair</strong> of P(1), P(2), ... P(N)
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<strong>if</strong> N < 2 <strong>then</strong>
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<strong>return</strong> ∞
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<strong>else</strong>
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minDistance ← |P(1) - P(2)|
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minPoints ← { P(1), P(2) }
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<strong>foreach</strong> i ∈ [1, N-1]
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<strong>foreach</strong> j ∈ [i+1, N]
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<strong>if</strong> |P(i) - P(j)| < minDistance <strong>then</strong>
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minDistance ← |P(i) - P(j)|
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minPoints ← { P(i), P(j) }
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<strong>endif</strong>
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<strong>endfor</strong>
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<strong>endfor</strong>
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<strong>return</strong> minDistance, minPoints
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<strong>endif</strong>
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</pre>
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```
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bruteForceClosestPair of P（1），P（2），... P（N）
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如果N <2那么
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返回∞
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其他
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minDistance←| P（1） - P（2）|
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minPoints←{P（1），P（2）}
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foreachi∈[1，N-1]
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foreachj∈[i + 1，N]
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if | P（i） - P（j）| <minDistance然后
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minDistance←| P（i） - P（j）|
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minPoints←{P（i），P（j）}
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万一
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ENDFOR
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ENDFOR
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return minDistance，minPoints
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万一
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```
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A better algorithm is based on the recursive divide and conquer approach, as explained also at [Wikipedia's Closest pair of points problem](https://en.wikipedia.org/wiki/Closest pair of points problem#Planar_case "wp: Closest pair of points problem#Planar_case"), which is `O(nlog(n))` a pseudo-code could be:
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</pre><p>一个更好的算法是基于递归分而治之的方法，正如<a href='https://en.wikipedia.org/wiki/Closest pair of points problem#Planar_case' title='wp：最近点的问题#Planar_case'>维基百科最近的一对点问题</a>所解释的那样，即O（n log n）;伪代码可以是： </p><pre>最近的（xP，yP）
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其中xP是P（1）.. P（N）按x坐标排序，和
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yP是P（1）.. P（N）按y坐标排序（升序）
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如果N≤3那么
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使用强力算法返回xP的最近点
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其他
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xL←xP点从1到⌈N/2⌉
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xR←xP点从⌈N/2⌉+ 1到N.
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xm←xP（⌈N/2⌉） <sub>x</sub>
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基←{P∈YP：P <sub>X≤XM}</sub>
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yR←{p∈yP：p <sub>x</sub> > xm}
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（dL，pairL）←nearestPair（xL，yL）
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（dR，pairR）←nearestRair（xR，yR）
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（dmin，pairMin）←（dR，pairR）
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如果dL <dR则
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（dmin，pairMin）←（dL，pairL）
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万一
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yS←{p∈yP：| xm - p <sub>x</sub> | <dmin}
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nS←yS中的点数
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（最近，最近的公里）←（dmin，pairMin）
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我从1到nS - 1
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k←i + 1
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而k≤nS和yS（k） <sub>y</sub> -yS（i） <sub>y</sub> <dmin
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if | yS（k） - yS（i）| <最接近的
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（最近，最近的公里）←（| yS（k） - yS（i）|，{yS（k），yS（i）}）
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万一
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k←k + 1
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ENDWHILE
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ENDFOR
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返回最近，最近的
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万一
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</pre>参考和进一步阅读： <a href='https://en.wikipedia.org/wiki/Closest pair of points problem' title='wp：最近点的问题'>最近的一对点问题</a> <a href='http://www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html' title='链接：http：//www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html'>最近的一对（麦吉尔）</a> <a href='http://www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf' title='链接：http：//www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf'>最近的一对（UCSB）</a> <a href='http://classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf' title='链接：http：//classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf'>最近的一对（WUStL）</a> <a href='http://www.cs.iupui.edu/~xkzou/teaching/CS580/Divide-and-conquer-closestPair.ppt' title='链接：http：//www.cs.iupui.edu/~xkzou/teaching/CS580/Divide-and-conquer-closestPair.ppt'>最近的一对（IUPUI）</a> <p>对于输入，期望参数是一个对象（点）数组，其中<code>x</code>和<code>y</code>成员设置为数字。对于输出，返回一个包含键的对象： <code>distance</code>和<code>pair</code>值对（即两对最近点的对）。 </p>
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<pre><strong>closestPair</strong> of (xP, yP)
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where xP is P(1) .. P(N) sorted by x coordinate, and
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yP is P(1) .. P(N) sorted by y coordinate (ascending order)
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<strong>if</strong> N ≤ 3 <strong>then</strong>
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<strong>return</strong> closest points of xP using brute-force algorithm
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<strong>else</strong>
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xL ← points of xP from 1 to ⌈N/2⌉
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xR ← points of xP from ⌈N/2⌉+1 to N
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xm ← xP(⌈N/2⌉)<sub>x</sub>
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yL ← { p ∈ yP : p<sub>x</sub> ≤ xm }
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yR ← { p ∈ yP : p<sub>x</sub> > xm }
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(dL, pairL) ← closestPair of (xL, yL)
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(dR, pairR) ← closestPair of (xR, yR)
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(dmin, pairMin) ← (dR, pairR)
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<strong>if</strong> dL < dR <strong>then</strong>
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(dmin, pairMin) ← (dL, pairL)
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<strong>endif</strong>
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yS ← { p ∈ yP : |xm - p<sub>x</sub>| < dmin }
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nS ← number of points in yS
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(closest, closestPair) ← (dmin, pairMin)
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<strong>for</strong> i <strong>from</strong> 1 <strong>to</strong> nS - 1
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k ← i + 1
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<strong>while</strong> k ≤ nS <strong>and</strong> yS(k)<sub>y</sub> - yS(i)<sub>y</sub> < dmin
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<strong>if</strong> |yS(k) - yS(i)| < closest <strong>then</strong>
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(closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
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<strong>endif</strong>
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k ← k + 1
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<strong>endwhile</strong>
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<strong>endfor</strong>
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<strong>return</strong> closest, closestPair
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<strong>endif</strong>
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</pre>
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For the input, expect the argument to be an array of objects (points) with `x` and `y` members set to numbers. For the output, return an object containing the key:value pairs for `distance` and `pair` (the pair of two closest points).
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**References and further readings:**
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<ul>
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<li><a href='https://en.wikipedia.org/wiki/Closest pair of points problem' title='wp: Closest pair of points problem' target='_blank'>Closest pair of points problem</a></li>
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<li><a href='https://www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html' target='_blank'>Closest Pair (McGill)</a></li>
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<li><a href='https://www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf' target='_blank'>Closest Pair (UCSB)</a></li>
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<li><a href='https://classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf' target='_blank'>Closest pair (WUStL)</a></li>
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</ul>
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# --hints--
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`getClosestPair`是一个函数。
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`getClosestPair` should be a function.
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```js
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assert(typeof getClosestPair === 'function');
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```
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距离应如下。
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Distance should be the following.
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```js
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assert.equal(getClosestPair(points1).distance, answer1.distance);
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```
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要点应如下。
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Points should be the following.
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```js
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assert.deepEqual(
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@ -89,13 +99,13 @@ assert.deepEqual(
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);
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```
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距离应如下。
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Distance should be the following.
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```js
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assert.equal(getClosestPair(points2).distance, answer2.distance);
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```
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要点应如下。
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Points should be the following.
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```js
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assert.deepEqual(
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