2018-10-10 22:03:03 +00:00
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---
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id: 5966f99c45e8976909a85575
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2020-12-16 07:37:30 +00:00
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title: 一周中的天
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2018-10-10 22:03:03 +00:00
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challengeType: 5
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videoUrl: ''
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---
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2020-12-16 07:37:30 +00:00
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# --description--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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<p>一家公司决定,每当圣诞节落在星期天,他们将给予他们的工人所有额外带薪假期,这样,在任何公共假期,工人将不必在下一周(12月25日到1月1日之间)工作。 </p><p>任务: </p><p>编写一个开始年份和结束年份的函数,并返回12月25日为星期日的所有年份的数组。 </p>
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --hints--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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`findXmasSunday`是一个函数。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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```js
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assert(typeof findXmasSunday === 'function');
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`findChristmasSunday(2000, 2100)`应该返回一个数组。
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert(typeof findXmasSunday(2000, 2100) === 'object');
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`findChristmasSunday(2008, 2121`应该回归[1977,1983,1988,1994,2005,2011,2016]
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert.deepEqual(findXmasSunday(1970, 2017), firstSolution);
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`findChristmasSunday(2008, 2121`应该返回[2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118]
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert.deepEqual(findXmasSunday(2008, 2121), secondSolution);
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2018-10-10 22:03:03 +00:00
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```
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2020-08-13 15:24:35 +00:00
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2020-12-16 07:37:30 +00:00
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# --solutions--
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