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Назарова Галина #9
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🍅 Пройдено тестов 10 из 24 |
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🍅 Пройдено тестов 10 из 24 |
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🍏 Пройдено тестов 24 из 24 |
lib.js
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| for (var j = 0; j < newLevelFriends.length; j++) { | ||
| friendsWithLevel.push( | ||
| { | ||
| friend: getFriendByName(friends, newLevelFriends[j]), |
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было бы круто избавиться от этой функции, например храня в newLevelFriends не только имена а сразу объект друга
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| friendsWithLevel.sort(function (friendOne, friendTwo) { | ||
| if (friendOne.level > friendTwo.level) { |
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можно красиво сделать через тернарник
lib.js
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| function Iterator(friends, filter) { | ||
| console.info(friends, filter); | ||
| if (!Filter.prototype.isPrototypeOf(filter)) { | ||
| throw new TypeError(); |
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лучше написать здесь ошибку
lib.js
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| if (!Filter.prototype.isPrototypeOf(filter)) { | ||
| throw new TypeError(); | ||
| } | ||
| this.friendsWithLevel = friendsLevel(friends).filter(function (friend) { |
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все методы нужно разместить в прототипах
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🍏 Пройдено тестов 24 из 24 |
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🚀 |
dzlk
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🍅
| }; | ||
| } | ||
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| MaleFilter.prototype = Object.create(Filter.prototype); |
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❗️ Свйоство (new MaleFilter()).contructor теперь указывает на Filter
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| Iterator.prototype.friendsLevel = function (friends) { | ||
| var bestFriends = friends.filter(function (friend) { | ||
| return friend.hasOwnProperty('best') && friend.best; |
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В принципе, можно заменить на Boolean(friend.best)
| for (var i = 0; i < friendsWithLevel.length; i++) { | ||
| var newLevelFriends = | ||
| friendsWithLevel[i].friend.friends.reduce(function (newLevelFunction, friendName) { | ||
| if (namesOfFriends.indexOf(friendName) === -1) { |
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❗️ Лучше держать объект имен, по нему скорость поиска значительно выше
| } | ||
| ); | ||
| namesOfFriends.push(newLevelFriends[j].name); | ||
| } |
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❗️ Обычно цикл for используется для предсказуемого числа итераций. Тут явно не тот случай.
Я бы предложил использовать очередь, из которой выталкивать по одному элементу. Имеется фор по friendsWithLevel
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| return (friendOne.friend.name > friendTwo.friend.name) ? 1 : -1; | ||
| }); |
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❗️ А можно без этой сортировки в конце обойтись?
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