困难的功能是LeapYear [关闭]

Question

我应该写一个函数 isLeapYear.t 应该接受一个毫秒数或日期字符串作为参数。此函数检查传递的参数是否为闰年。如果是闰年，函数应该返回一个字符串——‘“年”是闰年’，如果不是，应该返回字符串——‘“年”不是闰年’。我应该使用 Date 对象。我知道这不是我想要的工作

if (year % 400 == 0)
  leap = true;
else if (year % 100 == 0)
  leap = false;
else if (year % 4 == 0)
  leap = true;
else
  leap = false;

For Example isLeapYear('2020-01-01 00:00:00'); // =>  ‘2020 is a leap year’
isLeapYear('2020-01-01 00:00:00777'); // =>  ‘Invalid Date’
isLeapYear('2021-01-15 13:00:00');  // =>  ‘2021 is not a leap year’
isLeapYear('2200-01-15 13:00:00'); // =>  ‘2200 is not a leap year’
isLeapYear(1213131313135465656654564646542132132131); // =>  ‘Invalid Date’
isLeapYear(1213131313); ); // => ‘1970 is not a leap year’

Answer 1

我假设你想要 JS，尽管示例在上帝知道什么

const isLeapYearHelper = (year) => (year % 400 == 0 ? true : year % 100 == 0 ? false : year % 4 == 0 ? true : false)

const isLeapYear = (str) => {
const date = new Date(str)
const isValid = !isNaN(date.getTime())
return isValid ? isLeapYearHelper(date.getFullYear()) ? `${date.getFullYear()} is a leap year` : `${date.getFullYear()} is a not leap year` : 'Invalid Date'
}

console.log(isLeapYear('2020-01-01 00:00:00777'))
console.log(isLeapYear('2021-01-15 13:00:00'))
console.log(isLeapYear('2200-01-15 13:00:00'))
console.log(isLeapYear('2020-01-15 13:00:00'))
console.log(isLeapYear(1213131313135465656654564646542132132131))
console.log(isLeapYear(1213131313))