Translate algorithms #10

BigFaceMaster · 2018-07-22T03:42:55Z

翻译章节

《面试前期准备——算法》

变更文件

algorithms.json
algorithms.md

翻译进度

45%
0%

…nto translate

S1ngS1ng

所有的 markdown 语法都要改成 HTML 语法才行
建议采用 JavaScript 的语法举例
题目的描述，特别是有多个参数的时候，建议分开描述。先解释传入的每个参数的意义，然后再细致地描述题目逻辑。可以参考我的翻译：
- Diff Two Arrays
- Seek and destroy

S1ngS1ng · 2018-07-22T23:23:58Z

08-coding-interview-prep/algorithms.json

-        "Create a function that takes two or more arrays and returns an array of the <dfn>symmetric difference</dfn> (<code>&xutri;</code> or <code>&oplus;</code>) of the provided arrays.",
-        "Given two sets (for example set <code>A = {1, 2, 3}</code> and set <code>B = {2, 3, 4}</code>), the mathematical term \"symmetric difference\" of two sets is the set of elements which are in either of the two sets, but not in both (<code>A &xutri; B = C = {1, 4}</code>). For every additional symmetric difference you take (say on a set <code>D = {2, 3}</code>), you should get the set with elements which are in either of the two the sets but not both (<code>C &xutri; D = {1, 4} &xutri; {2, 3} = {1, 2, 3, 4}</code>). The resulting array must contain only unique values (<em>no duplicates</em>).",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：对称差 (Symmetric Difference)，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117' target='_blank'>集合</a> 的对称差是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合 A = { 1, 2, 3} 和 B = { 2, 3, 4} 的对称差为 A &xutri; B = C = {1,4},。 集合论中的这个运算相当于布尔逻辑中的异或运算。<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE' target='_blank'>百科</a>。",


建议把 对称差 的链接放到前面。

知识提要：<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE' target='_blank'>对称差</a>（Symmetric Difference）。

既然是 JavaScript 平台，建议用 [1, 2, 3] 来表示数组。下同

S1ngS1ng · 2018-07-22T23:26:10Z

08-coding-interview-prep/algorithms.json

-        "Given two sets (for example set <code>A = {1, 2, 3}</code> and set <code>B = {2, 3, 4}</code>), the mathematical term \"symmetric difference\" of two sets is the set of elements which are in either of the two sets, but not in both (<code>A &xutri; B = C = {1, 4}</code>). For every additional symmetric difference you take (say on a set <code>D = {2, 3}</code>), you should get the set with elements which are in either of the two the sets but not both (<code>C &xutri; D = {1, 4} &xutri; {2, 3} = {1, 2, 3, 4}</code>). The resulting array must contain only unique values (<em>no duplicates</em>).",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：对称差 (Symmetric Difference)，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117' target='_blank'>集合</a> 的对称差是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合 A = { 1, 2, 3} 和 B = { 2, 3, 4} 的对称差为 A &xutri; B = C = {1,4},。 集合论中的这个运算相当于布尔逻辑中的异或运算。<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE' target='_blank'>百科</a>。",
+        "创建一个函数sym，输入两个或两个以上的数组作为参数，然后返回值为 **对称差** 的数组",


然后返回值为 **对称差** 的数组 => 以数组的形式返回它们的对称差。

注意，这里用 markdown syntax 是不行的。如果要强调，也许要用 <em> 或者 <b> 这样的包裹起来

下同

S1ngS1ng · 2018-07-22T23:44:07Z

08-coding-interview-prep/algorithms.json

+        "知识提要：对称差 (Symmetric Difference)，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117' target='_blank'>集合</a> 的对称差是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合 A = { 1, 2, 3} 和 B = { 2, 3, 4} 的对称差为 A &xutri; B = C = {1,4},。 集合论中的这个运算相当于布尔逻辑中的异或运算。<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE' target='_blank'>百科</a>。",
+        "创建一个函数sym，输入两个或两个以上的数组作为参数，然后返回值为 **对称差** 的数组",
+        "思路：设定两个数组 (例如：<code>A = {1, 2, 3}</code>，<code>B = {2, 3, 4}</code>，)作为参数传入，返回对称差数组（<code>A &xutri; B = C = {1, 4}</code>），且数组元素具有唯一性（没有重复项）。",
+        "如果你遇到了困难，请点击 <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>帮助</a>。请尽量编写属于你自己的代码。尝试一下 **结对编程** 会有意料不到的惊喜哦。",


见群内讨论

以及： https://github.com/FreeCodeCampChina/challenges/pull/8/files#diff-7f72cbe6be2fd7054b07b7455775fde6R16

S1ngS1ng · 2018-07-22T23:55:11Z

08-coding-interview-prep/algorithms.json

-        "Compare and update the inventory stored in a 2D array against a second 2D array of a fresh delivery. Update the current existing inventory item quantities (in <code>arr1</code>). If an item cannot be found, add the new item and quantity into the inventory array. The returned inventory array should be in alphabetical order by item.",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "",
+        "比较和更新存储在 <a href='https://baike.baidu.com/item/%E4%BA%8C%E7%BB%B4%E6%95%B0%E7%BB%84' target='_blank'>2D数组 </a>中的库存，并将其与新交付的第二个2D数组进行比较。更新当前的库存项数量（arr1）。如果找不到项目，请将新项目和数量添加到库存数组中。返回的库存数组应该按照项目的字母顺序排列。",


2D数组 => 二维数组

建议更新题目描述，见 review comments

S1ngS1ng · 2018-07-23T04:40:41Z

@BigFaceMaster 另外，你添加了 .DS_Store 文件。请删除

…nto translate

S1ngS1ng

空格改一下。之前的 comments 也看一下，你还都没改。先别添加新的翻译内容了

S1ngS1ng · 2018-07-24T17:03:57Z

08-coding-interview-prep/algorithms.json

        },
        {
-          "text": "<code>permAlone(\"aab\")</code> should return 2.",
-          "testString": "assert.strictEqual(permAlone('aab'), 2,'<code>permAlone(\"aab\")</code> should return 2.');"
+          "text": "<code>permAlone(\"aab\")</code> 应该返回 2 。",


半角字符与全角标点之间不应该添加空格。见群内讨论

所以这里是 应该返回 2。
下同

S1ngS1ng · 2018-07-24T17:05:43Z

08-coding-interview-prep/algorithms.json

@@ -255,7 +255,7 @@
        "<table class=\"table\"><tr><th><strong>Index</strong></th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th></tr><tr><td>Value</td><td>7</td><td>9</td><td>11</td><td>13</td><td>15</td></tr></table>",
        "Below we'll take their corresponding indices and add them.",
        "7 + 13 = 20 &#8594; Indices 0 + 3 = 3<br>9 + 11 = 20 &#8594; Indices 1 + 2 = 3<br>3 + 3 = 6 &#8594 Return <code>6</code>",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514'target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "如果你遇到了困难，请点击 <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514'target='_blank'> 帮助 </a>。你可以找人 “结对编程”，但不要因此放弃思考。",


见群内讨论。行内元素是否加空格取决于里面的内容。
中文引号两边也是不需要加空格的

如果你遇到了困难，请点击<a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514'target='_blank'>帮助</a>。你可以找人“结对编程”，但不要因此放弃思考。

huluoyang · 2018-07-24T17:57:55Z

08-coding-interview-prep/algorithms.json

-        "Create a function that takes two or more arrays and returns an array of the <dfn>symmetric difference</dfn> (<code>&xutri;</code> or <code>&oplus;</code>) of the provided arrays.",
-        "Given two sets (for example set <code>A = {1, 2, 3}</code> and set <code>B = {2, 3, 4}</code>), the mathematical term \"symmetric difference\" of two sets is the set of elements which are in either of the two sets, but not in both (<code>A &xutri; B = C = {1, 4}</code>). For every additional symmetric difference you take (say on a set <code>D = {2, 3}</code>), you should get the set with elements which are in either of the two the sets but not both (<code>C &xutri; D = {1, 4} &xutri; {2, 3} = {1, 2, 3, 4}</code>). The resulting array must contain only unique values (<em>no duplicates</em>).",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：对称差 (Symmetric Difference)，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'> 集合 </a> 的对称差是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合 A = { 1, 2, 3} 和 B = { 2, 3, 4} 的对称差为 A &xutri; B = C = {1,4},。 集合论中的这个运算相当于布尔逻辑中的异或运算。<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE'target='_blank'> 百科 </a>。",


Symmetric Difference => 对称差分

@huluoyang https://zh.m.wikipedia.org/zh-hans/对称差

…nto translate

S1ngS1ng

@BigFaceMaster 之前的 comments 你还没有改，点开 Show outdated 看下。
比如数组的表示方式，markdown syntax 要删掉，以及<a>标签里面其实不需要加空格。
麻烦你更新下哈。内容方面应该没有问题了

…nto translate

BigFaceMaster · 2018-07-31T12:49:41Z

已经根据建议，更改并优化。

S1ngS1ng

LGTM. Thanks 👍

huluoyang · 2018-08-02T20:34:47Z

08-coding-interview-prep/algorithms.json

-        "Create a function that takes two or more arrays and returns an array of the <dfn>symmetric difference</dfn> (<code>&xutri;</code> or <code>&oplus;</code>) of the provided arrays.",
-        "Given two sets (for example set <code>A = {1, 2, 3}</code> and set <code>B = {2, 3, 4}</code>), the mathematical term \"symmetric difference\" of two sets is the set of elements which are in either of the two sets, but not in both (<code>A &xutri; B = C = {1, 4}</code>). For every additional symmetric difference you take (say on a set <code>D = {2, 3}</code>), you should get the set with elements which are in either of the two the sets but not both (<code>C &xutri; D = {1, 4} &xutri; {2, 3} = {1, 2, 3, 4}</code>). The resulting array must contain only unique values (<em>no duplicates</em>).",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE'target='_blank'>对称差 (Symmetric Difference)</a>，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'>集合</a> 的对称差分是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合<code>let A = [ 1, 2, 3]</code>和<code>let B = [ 2, 3, 4]</code>的对称差分为<code>A &xutri; B = C = [ 1, 4]</code>,。 集合论中的这个运算相当于布尔逻辑中的异或运算。",


知识提要👍
对称差分 => 对等差分

huluoyang · 2018-08-02T20:36:26Z

08-coding-interview-prep/algorithms.json

-        "Compare and update the inventory stored in a 2D array against a second 2D array of a fresh delivery. Update the current existing inventory item quantities (in <code>arr1</code>). If an item cannot be found, add the new item and quantity into the inventory array. The returned inventory array should be in alphabetical order by item.",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "",
+        "比较和更新存储在<a href='https://baike.baidu.com/item/%E4%BA%8C%E7%BB%B4%E6%95%B0%E7%BB%84'target='_blank'>二维数组</a>中的库存，并将其与新交付的第二个二维数组进行比较。更新当前的库存项数量（arr1）。如果找不到项目，请将新项目和数量添加到库存数组中。返回的库存数组应该按照项目的字母顺序排列。",


符号使用混乱，行文不够流畅。
请重新翻译。

S1ngS1ng

@BigFaceMaster 内容上没啥大问题。但其他部分，不翻译的/不需要改的部分就不要改，保持和原文一样就行，否则后续很难做 diff。

格式问题记得参考 style-guide

S1ngS1ng · 2018-08-06T03:29:31Z

08-coding-interview-prep/algorithms.json

-        "Create a function that takes two or more arrays and returns an array of the <dfn>symmetric difference</dfn> (<code>&xutri;</code> or <code>&oplus;</code>) of the provided arrays.",
-        "Given two sets (for example set <code>A = {1, 2, 3}</code> and set <code>B = {2, 3, 4}</code>), the mathematical term \"symmetric difference\" of two sets is the set of elements which are in either of the two sets, but not in both (<code>A &xutri; B = C = {1, 4}</code>). For every additional symmetric difference you take (say on a set <code>D = {2, 3}</code>), you should get the set with elements which are in either of the two the sets but not both (<code>C &xutri; D = {1, 4} &xutri; {2, 3} = {1, 2, 3, 4}</code>). The resulting array must contain only unique values (<em>no duplicates</em>).",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE'target='_blank'>对称差 (Symmetric Difference)</a>，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'>集合</a> 的对称差分是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合<code>let A = [ 1, 2, 3]</code>和<code>let B = [ 2, 3, 4]</code>的对称差分为<code>A &xutri; B = C = [ 1, 4]</code>,。 集合论中的这个运算相当于布尔逻辑中的异或运算。",


两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'>集合</a> 的对称差分是只属于其中一个集合，

由于 <a> 里面和外面都是全角，因此 <a> 两边不需要加空格。

最后哪个 </code> 之后多了一个半角逗号。

S1ngS1ng · 2018-08-06T03:30:32Z

08-coding-interview-prep/algorithms.json

-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "知识提要：<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE'target='_blank'>对称差 (Symmetric Difference)</a>，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'>集合</a> 的对称差分是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合<code>let A = [ 1, 2, 3]</code>和<code>let B = [ 2, 3, 4]</code>的对称差分为<code>A &xutri; B = C = [ 1, 4]</code>,。 集合论中的这个运算相当于布尔逻辑中的异或运算。",
+        "创建一个函数 sym，输入两个或两个以上的数组作为参数，然后返回值为<em>对称差分</em>的数组",
+        "思路：设定两个数组 (例如：<code>let A = [1, 2, 3]</code>，<code>let B = [2, 3, 4]</code>，)作为参数传入，返回对称差分数组（<code>A &xutri; B = C = [1, 4]</code>），且数组元素具有唯一性（没有重复项）。",


第一个括号里面，） 之前不应该加逗号

且数组元素具有唯一性（没有重复项）。 =>
且数组中没有重复项。

S1ngS1ng · 2018-08-06T03:32:34Z

08-coding-interview-prep/algorithms.json

+        "知识提要：<a href='https://baike.baidu.com/item/%E5%AF%B9%E7%A7%B0%E5%B7%AE'target='_blank'>对称差 (Symmetric Difference)</a>，数学上，两个 <a href='https://baike.baidu.com/item/%E9%9B%86%E5%90%88/2908117'target='_blank'>集合</a> 的对称差分是只属于其中一个集合，而不属于另一个集合的元素组成的集合，例如：集合<code>let A = [ 1, 2, 3]</code>和<code>let B = [ 2, 3, 4]</code>的对称差分为<code>A &xutri; B = C = [ 1, 4]</code>,。 集合论中的这个运算相当于布尔逻辑中的异或运算。",
+        "创建一个函数 sym，输入两个或两个以上的数组作为参数，然后返回值为<em>对称差分</em>的数组",
+        "思路：设定两个数组 (例如：<code>let A = [1, 2, 3]</code>，<code>let B = [2, 3, 4]</code>，)作为参数传入，返回对称差分数组（<code>A &xutri; B = C = [1, 4]</code>），且数组元素具有唯一性（没有重复项）。",
+        "如果你遇到了困难，请点击 <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514'target='_blank'>帮助</a>。你可以找人“结对编程”，但不要因此放弃思考。",


同理，这里的 “帮助” 之前也不需要加空格。
但引号前是要加的。参考 style-guide

S1ngS1ng · 2018-08-06T03:33:42Z

08-coding-interview-prep/algorithms.json

@@ -99,36 +100,36 @@
      "id": "a56138aff60341a09ed6c480",
      "title": "Inventory Update",
      "description": [
-        "Compare and update the inventory stored in a 2D array against a second 2D array of a fresh delivery. Update the current existing inventory item quantities (in <code>arr1</code>). If an item cannot be found, add the new item and quantity into the inventory array. The returned inventory array should be in alphabetical order by item.",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "创建一个<a href='https://baike.baidu.com/item/%E4%BA%8C%E7%BB%B4%E6%95%B0%E7%BB%84'target='_blank'>二维数组</a>，比较并更新存储在<code>二维数组</code>中的”库存“元素，然后并将其与新产生的第二个二维数组进行对比，更新当前的”库存“项的数量（<code>arr1</code>），如果找不到这个对比对象，那么将新的对象和数量添加到“库存”数组中。注意：返回的“库存”数组应该是按照数组元素的首字母顺序排序",


这里的第一组引号好像用反了。。另外记得加空格

S1ngS1ng · 2018-08-06T03:34:12Z

08-coding-interview-prep/algorithms.json

-        "For example, <code>aab</code> should return 2 because it has 6 total permutations (<code>aab</code>, <code>aab</code>, <code>aba</code>, <code>aba</code>, <code>baa</code>, <code>baa</code>), but only 2 of them (<code>aba</code> and <code>aba</code>) don't have the same letter (in this case <code>a</code>) repeating.",
-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "把一个字符串中的字符重新排列生成新的字符串，返回新生成的字符串里没有连续重复字符的字符串个数。连续重复只以单个字符为准。",
+        "例如：<code>aab</code>应该返回 2， 因为它总共有 6 种排列方式： <code>aab</code>， <code>aab</code>， <code>aba</code>， <code>aba</code>， <code>baa</code>， <code>baa</code>，但是其中只有 2 个没有连续重复的字符（ 字符 a 是本例中的重复字符 ）：<code>aba</code>，<code>aba</code>",


全角符号之后不加空格。

S1ngS1ng · 2018-08-06T03:34:44Z

08-coding-interview-prep/algorithms.json

-        "Remember to use <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514' target='_blank'>Read-Search-Ask</a> if you get stuck. Try to pair program. Write your own code."
+        "把一个字符串中的字符重新排列生成新的字符串，返回新生成的字符串里没有连续重复字符的字符串个数。连续重复只以单个字符为准。",
+        "例如：<code>aab</code>应该返回 2， 因为它总共有 6 种排列方式： <code>aab</code>， <code>aab</code>， <code>aba</code>， <code>aba</code>， <code>baa</code>， <code>baa</code>，但是其中只有 2 个没有连续重复的字符（ 字符 a 是本例中的重复字符 ）：<code>aba</code>，<code>aba</code>",
+        "如果你遇到了困难，请点击 <a href='http://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck/19514'target='_blank'> 帮助 </a>。你可以找人” 结对编程 “，但不要因此放弃思考。"


你可以找人” 结对编程 “

这里引号用的也不对

S1ngS1ng · 2018-08-06T03:35:20Z

08-coding-interview-prep/algorithms.json

      ],
      "solutions": [
-        "function permAlone(str) {\n  return permutor(str).filter(function(perm) {\n    return !perm.match(/(.)\\1/g);\n  }).length;\n}\n\nfunction permutor(str) {\n  // http://staff.roguecc.edu/JMiller/JavaScript/permute.html\n  //permArr: Global array which holds the list of permutations\n  //usedChars: Global utility array which holds a list of \"currently-in-use\" characters\n  var permArr = [], usedChars = [];\n  function permute(input) {\n    //convert input into a char array (one element for each character)\n    var i, ch, chars = input.split(\"\");\n    for (i = 0; i < chars.length; i++) {\n      //get and remove character at index \"i\" from char array\n      ch = chars.splice(i, 1);\n      //add removed character to the end of used characters\n      usedChars.push(ch);\n      //when there are no more characters left in char array to add, add used chars to list of permutations\n      if (chars.length === 0) permArr[permArr.length] = usedChars.join(\"\");\n      //send characters (minus the removed one from above) from char array to be permuted\n      permute(chars.join(\"\"));\n      //add removed character back into char array in original position\n      chars.splice(i, 0, ch);\n      //remove the last character used off the end of used characters array\n      usedChars.pop();\n    }\n  }\n  permute(str);\n  return permArr;\n}\n\npermAlone('aab');\n"
+        "function permAlone(str) {\n  return permutor(str).filter(function(perm) {\n    return !perm.match(/(.)\\1/g);\n  }).length;\n}\n\nfunction permutor(str) {\n  // http://staff.roguecc.edu/JMiller/JavaScript/permute.html\n  //permArr: 全局数组：包含排列的列表 \ n  //usedChars: 全局实用数组，其中一个列表包含 \"currently-in-use\"字符  var permArr = [], usedChars = [];\n  function permute(input) {\n    // 将输入参数转换成一个字符串数组 （每个字符都有一个元素）\n    var i, ch, chars = input.split(\"\");\n    for (i = 0; i < chars.length; i++) {\n      // 从字符数组中获取并删除索引为\"i\"的字符 \n      ch = chars.splice(i, 1);\n      // 将删除的字符增加的使用字符的末尾 \n      usedChars.push(ch);\n      // 当char数组中没有其他字符添加时，添加使用的chars列表来排列排列 \n      if (chars.length === 0) permArr[permArr.length] = usedChars.join(\"\");\n      // 从char数组发送字符（减去上面删除的字符）进行置换 \n      permute(chars.join(\"\"));\n      // 将删除的数组添加回char数组中的原位置 \n      chars.splice(i, 0, ch);\n      //删除使用过的字符数组末尾使用的最后一个字符 \n      usedChars.pop();\n    }\n  }\n  permute(str);\n  return permArr;\n}\n\npermAlone('aab');\n"


这个 solutions 是不需要翻译的。 @huluoyang 确认下？我记得这个只是用来执行判断的

S1ngS1ng · 2018-08-06T03:36:00Z

08-coding-interview-prep/algorithms.json

@@ -233,7 +234,7 @@
          "name": "index",
          "contents": [
            "function permAlone(str) {",
-            "  return str;",
+            "return str;",


这里的缩进不要移除，这个是用来在编辑器中显示代码的，去掉之后缩进就没了。

这里怕是插件的锅哦。。。

S1ngS1ng · 2018-08-06T03:36:52Z

08-coding-interview-prep/algorithms.json

@@ -351,8 +352,8 @@
          "head": [],
          "tail": [
            "function isSorted(arr) {",
-            "  var check = (i) => (i == arr.length - 1) ? true : (arr[i] > arr[i + 1]) ? false : check(i + 1);",
-            "  return check(0);",
+            "var check = (i) => (i == arr.length - 1) ? true : (arr[i] > arr[i + 1]) ? false : check(i + 1);",


这里改了应该没事儿，但最好不要动，否则后续更新的时候会影响 diff 的判断

插件。。。

S1ngS1ng · 2018-08-06T20:58:35Z

@BigFaceMaster 好 👍 把前面 comment 里提到的改下就好啦，没别的问题

… translate) Merge Request: functional-programming.json complete Created By: @黄雨纯 Accepted By: @黄雨纯 URL: https://freecodecamp.coding.net/p/challenges/git/merge/10

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…nto translate

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No Repeats Please

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huluoyang changed the title ~~完成“面试前期准备-算法-Symmetric Difference”~~ Translate algorithms Jul 24, 2018

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Merge branch 'translate' of github.com:FreeCodeCampChina/challenges i…

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Merge branch 'translate' of github.com:FreeCodeCampChina/challenges i…

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Translate algorithms #10

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BigFaceMaster commented Jul 22, 2018 • edited Loading

翻译章节

变更文件

翻译进度

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