Hackerrank:打破记录
Hackerrank:打破记录
提问于 2020-10-22 14:38:47
我正在学习Clojure，并且是一名排名较高的n00b，试图从书籍和在线教程中学习(但有时我担心我正在养成坏习惯，或者至少不是所有的好习惯)。在练习中，我在Hackerrank上做了打破纪录问题。
TL;DR问题描述：
For一个分数列表(按历史顺序排列)，计算以前最好的分数被超过的次数，以及以前最差的分数被低估的次数。
在迭代语言中，使用只需遍历列表非常容易；但是Clojure不执行迭代，我决定在构造(尾部)递归解决方案时解决这个问题。为了让自己更容易，我首先用Java做了一个递归解决方案，然后我翻译了这个解决方案。总之，这是一个非常简单的递归函数，而不涉及火箭手术。
显然，我的代码可以工作，如包含的单元测试所示。不过，我所关注的事项如下：
- 当放在Java代码旁边时，两者看起来非常相似。我是遵循“熟能生巧”的Clojure编程，还是只是笨拙的“逐字音译”？
- 通过使用不同的Clojure构造，是否有更紧凑和/或更容易理解的区域？
您的关键输入将受到高度赞赏--关于单元测试的<#>including，因为这可以说是编程的一个重要部分，我想并行学习和实践。
代码：
(ns hackerrank.breaking-records
(:require [clojure.test :refer :all]))
(defrecord Record [min max countworse countbetter])
(defn recalc-record [rec newscore]
(Record.
(min newscore (:min rec))
(max newscore (:max rec))
(+ (:countworse rec) (if (> (:min rec) newscore) 1 0))
(+ (:countbetter rec) (if (< (:max rec) newscore) 1 0))))
(defn accumulate [curr-record remaining-scores]
(if (nil? (second remaining-scores))
curr-record
(recur (recalc-record curr-record (second remaining-scores)) (rest remaining-scores)))
)
(defn breaking-records [scores]
(let [result (accumulate (Record. (first scores) (first scores) 0 0) scores)]
(list (:countbetter result) (:countworse result))))
(deftest test-records
(testing "edge cases"
(is (= '(0 0) (breaking-records '())) "no games played yet")
(is (= '(0 0) (breaking-records '(5))) "single game"))
(testing "hackerrank examples"
(is (= '(2 4) (breaking-records '(10 5 20 20 4 5 2 25 1))))
(is (= '(4 0) (breaking-records '(3 4 21 36 10 28 35 5 24 42)))))
)回答 1
Code Review用户
回答已采纳
发布于 2020-10-22 19:14:10
我重写了您的解决方案，以使用更典型的Clojure特性。当您正在遍历数据并需要跟踪累积状态时，很难超过loop/recur。第一个例子是：
(ns tst.demo.core
(:use clojure.test))
(defn breaking-records
[scores]
; this loop has 5 variables. Init all of them
(loop [low (first scores)
high (first scores)
nworse 0
nbetter 0
score-pairs (partition 2 1 scores)]
(if (empty? score-pairs)
{:nbetter nbetter :nworse nworse}
(let [curr-score-pair (first score-pairs)
new-score (second curr-score-pair)]
; start the next iteration with modified versions of the 5 loop vars
(recur
(min new-score low)
(max new-score high)
(if (< new-score low)
(inc nworse)
nworse)
(if (< high new-score)
(inc nbetter)
nbetter)
(rest score-pairs))))))和单元测试：
(deftest test-records
(testing "edge cases"
(is (= (breaking-records []) {:nbetter 0 :nworse 0}) "no games played yet")
(is (= (breaking-records [5]) {:nbetter 0 :nworse 0}) "single game"))
(testing "hackerrank examples"
(is (= (breaking-records [10 5 20 20 4 5 2 25 1]) {:nbetter 2 :nworse 4}))
(is (= (breaking-records [3 4 21 36 10 28 35 5 24 42]) {:nbetter 4 :nworse 0}))))
; ***** NOTE: it's much easier to use vectors like [1 2 3] instead of a quoted list `(1 2 3)请见这份文件清单 (尤指)。Clojure CheatSheet。此外，模板项目作为一个整体展示了我是如何构造事物的。:)
帮助最大的函数是partition。看医生。
轻微重构
您可以通过使用更专门的函数(如reduce和cond-> )来简化它，并使其更加紧凑。此版本使用映射保持状态，使用reduce执行循环：
(defn breaking-records
[scores]
(let [state-init {:low (first scores)
:high (first scores)
:nworse 0
:nbetter 0}
accum-stats-fn (fn [state score-pair]
; Use map destructuring to pull out the 4 state variables
(let [{:keys [low high nworse nbetter]} state
new-score (second score-pair)
state-new {:low (min new-score low)
:high (max new-score high)
:nworse (cond-> nworse
(< new-score low) (inc))
:nbetter (cond-> nbetter
(< high new-score) (inc))}]
state-new))
state-final (reduce accum-stats-fn
state-init
(partition 2 1 scores))
result (select-keys state-final [:nworse :nbetter])]
result))页面原文内容由Code Review提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://codereview.stackexchange.com/questions/251024
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