ee1e8abd87
* 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>
145 lines
3.8 KiB
Markdown
145 lines
3.8 KiB
Markdown
---
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id: a3f503de51cf954ede28891d
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title: 找到对称差异
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challengeType: 5
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videoUrl: ''
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dashedName: find-the-symmetric-difference
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---
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# --description--
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创建一个带有两个或更多数组的函数，并返回所提供数组的<dfn>对称差</dfn> （ `△`或`⊕` ）数组。给定两个集合（例如集合`A = {1, 2, 3}`并且集合`B = {2, 3, 4}` ），两个集合的数学术语“对称差异”是在任一集合中的元素集合。两组，但两者都没有（ `A △ B = C = {1, 4}` ）。对于你所采取的每一个额外的对称差异（比如在集合`D = {2, 3}` ），你应该得到具有两个集合中的任何一个但不是两个集合的元素的集合（ `C △ D = {1, 4} △ {2, 3} = {1, 2, 3, 4}` ）。结果数组必须仅包含唯一值（ *不重复* ）。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。尝试配对程序。编写自己的代码。
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# --hints--
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`sym([1, 2, 3], [5, 2, 1, 4])`应返回`[3, 4, 5]` 。
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```js
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assert.sameMembers(sym([1, 2, 3], [5, 2, 1, 4]), [3, 4, 5]);
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```
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`sym([1, 2, 3], [5, 2, 1, 4])`应仅包含三个元素。
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```js
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assert.equal(sym([1, 2, 3], [5, 2, 1, 4]).length, 3);
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```
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`sym([1, 2, 3, 3], [5, 2, 1, 4])`应该返回`[3, 4, 5]` 。
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```js
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assert.sameMembers(sym([1, 2, 3, 3], [5, 2, 1, 4]), [3, 4, 5]);
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```
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`sym([1, 2, 3, 3], [5, 2, 1, 4])`应仅包含三个元素。
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```js
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assert.equal(sym([1, 2, 3, 3], [5, 2, 1, 4]).length, 3);
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```
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`sym([1, 2, 3], [5, 2, 1, 4, 5])`应返回`[3, 4, 5]` 。
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```js
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assert.sameMembers(sym([1, 2, 3], [5, 2, 1, 4, 5]), [3, 4, 5]);
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```
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`sym([1, 2, 3], [5, 2, 1, 4, 5])`应仅包含三个元素。
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```js
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assert.equal(sym([1, 2, 3], [5, 2, 1, 4, 5]).length, 3);
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```
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`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])`应该返回`[1, 4, 5]`
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```js
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assert.sameMembers(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]), [1, 4, 5]);
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```
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`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])`应仅包含三个元素。
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```js
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assert.equal(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]).length, 3);
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```
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`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])`应该返回`[1, 4, 5]` 。
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```js
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assert.sameMembers(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]), [1, 4, 5]);
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```
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`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])`应仅包含三个元素。
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```js
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assert.equal(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]).length, 3);
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```
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`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])`应该返回`[2, 3, 4, 6, 7]` `sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])` `[2, 3, 4, 6, 7]` 。
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```js
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assert.sameMembers(
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sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]),
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[2, 3, 4, 6, 7]
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);
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```
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`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])`应仅包含五个元素。
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```js
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assert.equal(
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sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3]).length,
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5
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);
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```
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`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])`应该返回`[1, 2, 4, 5, 6, 7, 8, 9]` 。
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```js
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assert.sameMembers(
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sym(
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[3, 3, 3, 2, 5],
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[2, 1, 5, 7],
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[3, 4, 6, 6],
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[1, 2, 3],
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[5, 3, 9, 8],
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[1]
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),
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[1, 2, 4, 5, 6, 7, 8, 9]
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);
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```
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`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])`应该只包含八个元素。
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```js
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assert.equal(
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sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])
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.length,
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8
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);
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```
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# --seed--
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## --seed-contents--
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```js
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function sym(args) {
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return args;
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}
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sym([1, 2, 3], [5, 2, 1, 4]);
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```
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# --solutions--
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```js
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function sym() {
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var arrays = [].slice.call(arguments);
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return arrays.reduce(function (symDiff, arr) {
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return symDiff.concat(arr).filter(function (val, idx, theArr) {
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return theArr.indexOf(val) === idx
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&& (symDiff.indexOf(val) === -1 || arr.indexOf(val) === -1);
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});
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});
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}
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sym([1, 2, 3], [5, 2, 1, 4]);
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```
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