* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
10 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
5956795bc9e2c415eb244de1 | 哈希加入 | 5 | hash-join |
--description--
内连接是一种操作,它根据匹配的列值将两个数据表组合到一个表中。实现此操作的最简单方法是嵌套循环连接算法,但更可扩展的替代方法是散列连接算法。
实现“散列连接”算法,并演示它通过下面列出的测试用例。
您应该将表表示为在编程语言中感觉自然的数据结构。
“散列连接”算法包含两个步骤:
哈希阶段:从两个表中的一个表创建一个多图,从每个连接列值映射到包含它的所有行。多图必须支持基于散列的查找,它比简单的线性搜索更好地扩展,因为这是该算法的重点。理想情况下,我们应该为较小的表创建多图,从而最小化其创建时间和内存大小。加入阶段:扫描另一个表,通过查看之前创建的多图来查找匹配的行。在伪代码中,算法可以表示如下:
让A =第一个输入表(或理想情况下,更大的输入表) 设B =第二个输入表(或理想情况下,较小的输入表) 令j A =表A的连接列ID 令j B =表B的连接列ID 让M B =一个多图,用于从单个值映射到表B的多行(从空白开始) 让C =输出表(从空开始) 对于表B中的每一行b: 将b放在密钥b(j B )下的多映射M B中 对于表A中的每一行a: 对于a(j A )项下多图M B中的每一行b: 设c =第a行和第b行的串联 将行c放在表C中测试用例
输入
|
产量
A.Age | 一个名字 | B.Character | B.Nemesis |
---|---|---|---|
27 | 约拿 | 约拿 | 鲸鱼 |
27 | 约拿 | 约拿 | 蜘蛛 |
18 | 艾伦 | 艾伦 | 鬼 |
18 | 艾伦 | 艾伦 | 植物大战僵尸 |
28 | 荣耀 | 荣耀 | 巴菲 |
28 | 艾伦 | 艾伦 | 鬼 |
28 | 艾伦 | 艾伦 | 植物大战僵尸 |
输出表中行的顺序并不重要。
如果你使用数字索引数组来表示表行(而不是按名称引用列),你可以用[[27, "Jonah"], ["Jonah", "Whales"]]
的形式表示输出行。 。
--hints--
hashJoin
是一个函数。
assert(typeof hashJoin === 'function');
hashJoin([{ age: 27, name: "Jonah" }, { age: 18, name: "Alan" }, { age: 28, name: "Glory" }, { age: 18, name: "Popeye" }, { age: 28, name: "Alan" }], [{ character: "Jonah", nemesis: "Whales" }, { character: "Jonah", nemesis: "Spiders" }, { character: "Alan", nemesis: "Ghosts" }, { character:"Alan", nemesis: "Zombies" }, { character: "Glory", nemesis: "Buffy" }, { character: "Bob", nemesis: "foo" }])
应该返回[{"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Whales"}, {"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Spiders"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}, {"A_age": 28,"A_name": "Glory", "B_character": "Glory", "B_nemesis": "Buffy"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}]
assert.deepEqual(hashJoin(hash1, hash2), res);
--seed--
--after-user-code--
const hash1 = [
{ age: 27, name: 'Jonah' },
{ age: 18, name: 'Alan' },
{ age: 28, name: 'Glory' },
{ age: 18, name: 'Popeye' },
{ age: 28, name: 'Alan' }
];
const hash2 = [
{ character: 'Jonah', nemesis: 'Whales' },
{ character: 'Jonah', nemesis: 'Spiders' },
{ character: 'Alan', nemesis: 'Ghosts' },
{ character: 'Alan', nemesis: 'Zombies' },
{ character: 'Glory', nemesis: 'Buffy' },
{ character: 'Bob', nemesis: 'foo' }
];
const res = [
{ A_age: 27, A_name: 'Jonah', B_character: 'Jonah', B_nemesis: 'Whales' },
{ A_age: 27, A_name: 'Jonah', B_character: 'Jonah', B_nemesis: 'Spiders' },
{ A_age: 18, A_name: 'Alan', B_character: 'Alan', B_nemesis: 'Ghosts' },
{ A_age: 18, A_name: 'Alan', B_character: 'Alan', B_nemesis: 'Zombies' },
{ A_age: 28, A_name: 'Glory', B_character: 'Glory', B_nemesis: 'Buffy' },
{ A_age: 28, A_name: 'Alan', B_character: 'Alan', B_nemesis: 'Ghosts' },
{ A_age: 28, A_name: 'Alan', B_character: 'Alan', B_nemesis: 'Zombies' }
];
const bench1 = [{ name: 'u2v7v', num: 1 }, { name: 'n53c8', num: 10 }, { name: 'oysce', num: 9 }, { name: '0mto2s', num: 1 }, { name: 'vkh5id', num: 4 }, { name: '5od0cf', num: 8 }, { name: 'uuulue', num: 10 }, { name: '3rgsbi', num: 9 }, { name: 'kccv35r', num: 4 }, { name: '80un74', num: 9 }, { name: 'h4pp3', num: 6 }, { name: '51bit', num: 7 }, { name: 'j9ndf', num: 8 }, { name: 'vf3u1', num: 10 }, { name: 'g0bw0om', num: 10 }, { name: 'j031x', num: 7 }, { name: 'ij3asc', num: 9 }, { name: 'byv83y', num: 8 }, { name: 'bjzp4k', num: 4 }, { name: 'f3kbnm', num: 10 }];
const bench2 = [{ friend: 'o8b', num: 8 }, { friend: 'ye', num: 2 }, { friend: '32i', num: 5 }, { friend: 'uz', num: 3 }, { friend: 'a5k', num: 4 }, { friend: 'uad', num: 7 }, { friend: '3w5', num: 10 }, { friend: 'vw', num: 10 }, { friend: 'ah', num: 4 }, { friend: 'qv', num: 7 }, { friend: 'ozv', num: 2 }, { friend: '9ri', num: 10 }, { friend: '7nu', num: 4 }, { friend: 'w3', num: 9 }, { friend: 'tgp', num: 8 }, { friend: 'ibs', num: 1 }, { friend: 'ss7', num: 6 }, { friend: 'g44', num: 9 }, { friend: 'tab', num: 9 }, { friend: 'zem', num: 10 }];
--seed-contents--
function hashJoin(hash1, hash2) {
return [];
}
--solutions--
function hashJoin(hash1, hash2) {
const hJoin = (tblA, tblB, strJoin) => {
const [jA, jB] = strJoin.split('=');
const M = tblB.reduce((a, x) => {
const id = x[jB];
return (
a[id] ? a[id].push(x) : (a[id] = [x]),
a
);
}, {});
return tblA.reduce((a, x) => {
const match = M[x[jA]];
return match ? (
a.concat(match.map(row => dictConcat(x, row)))
) : a;
}, []);
};
const dictConcat = (dctA, dctB) => {
const ok = Object.keys;
return ok(dctB).reduce(
(a, k) => (a[`B_${k}`] = dctB[k]) && a,
ok(dctA).reduce(
(a, k) => (a[`A_${k}`] = dctA[k]) && a, {}
)
);
};
return hJoin(hash1, hash2, 'name=character');
}