javascript - 编写一个函数，该函数接受一个整数数组和一个“偶数”或“奇数”字符串

Question

我正在处理 javascript 中的一个问题，我应该编写一个函数，该函数接收一个整数数组和一个“偶数”或“奇数”的字符串。该函数将计算连续出现 4 个偶数或 4 个奇数的次数。

例如：

quadruples([3,2,2,4,8,5], 'even')  // 1
quadruples([2,4,6,8,10,5], 'even')  // 2
quadruples([2,4,6,8,10,5], 'odd')  // 0

到目前为止，这是我所在的位置：

function quadruples(givenArray, evenOrOdd) {
  let arr = []
  if(evenOrOdd == 'even') {
    if( i = 0; i < givenArray.length; i++) {
      
    }
};

我想我需要运行一个 for 循环，然后使用 % 运算符，但我被困在从这里去哪里。

任何帮助表示赞赏！

标签： javascriptarraysfor-loopmodulus

Answer 1

您需要使用局部和全局变量对此进行动态编程： [2, 4, 6, 8, 10, 5]

• 2 - 偶数，count 为 1，totalCount 为 0
• 4 - 偶数，计数为 2，totalCount 为 0
• 6 - 偶数，计数为 3，totalCount 为 0
• 8 - 偶数，count 为 4，totalCount 为 0
• 10 - 偶数，计数为 5，totalCount 为 0
• 5 - 奇数，计数为 5，将 totalCount 增加 5 - 4 + 1 = 2，将计数重置为 0

const quadruples = (givenArray, evenOrOdd) => {
  // never hardcode `magic numbers`, create constants for them
  const sequenceLength = 4

  // based on evenOrOdd calculating what the division by 2
  // will be if it is even, then 0, if it is odd, then 1
  const rest = evenOrOdd === 'even' ? 0 : 1

  // this will hold the total count of quadruples
  let totalCount = 0

  // this is the local count of contiguous elements
  let count = 0

  // looping over the array
  for (let i = 0; i <= givenArray.length; i += 1) {
    const el = givenArray[i]

    // if the element is not what we want
    if (i === givenArray.length || el % 2 !== rest) {
      // if the count is 4 or more, we add to totalCount the count
      // minus 4 and plus 1, meaning that if we have 4, it's 1 quadruple,
      // if it is 5, then it's 2 quadruples, etc.
      // Otherwise (count is less than 4) we add 0 (nothing)
      totalCount += count >= sequenceLength ? count - sequenceLength + 1 : 0

      // resetting the count to zero as we encountered the opposite
      // of what we are looking for (even/odd)
      count = 0

      // if the element is what we need (even or odd)
    } else {
      // increasing the count of how many we've seen by far
      count += 1
    }
  }

  // returning totalCount of quadruples
  return totalCount
}

console.log(quadruples([1, 3, 5, 7, 9, 11], 'odd')) // 3
console.log(quadruples([3, 2, 2, 4, 8, 5], 'even')) // 1
console.log(quadruples([2, 4, 6, 8, 10, 5], 'even')) // 2
console.log(quadruples([2, 4, 6, 8, 10, 5], 'odd')) // 0