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仅添加奇数或偶数 .

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-4

我想创建一个函数，当给定一个数字数组和一个条件（例如奇数或偶数）时，匹配该条件的数字被加在一起 . 如果没有值与条件匹配，则应返回0 .

如果没有Array.prototype.filter（），我怎么能这样做 - 我的教科书告诉我没有这种方法就可以做到这一点 .

如果我应该使用什么陈述？

console.log(conditionalSum([1, 2, 3, 4, 5], "even"));        => 6
console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));         => 9
console.log(conditionalSum([13, 88, 12, 44, 99], "even"));   => 144
console.log(conditionalSum([], "odd"));                      => 0

谢谢 .

javascript arrays

6 回答

  • 3

    您可以通过选择正确的回调来减少数组 .

    function conditionalSum(array, parts) {
    return array.reduce(
        parts === 'even' 
            ? (s, v) => s + (!(v % 2) && v)
            : (s, v) => s + (v % 2 && v),
        0
   );
}

console.log(conditionalSum([1, 2, 3, 4, 5], "even"));      //   6
console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));       //   9
console.log(conditionalSum([13, 88, 12, 44, 99], "even")); // 144
console.log(conditionalSum([], "odd"));                    //   0
    回复于
  • -2
    function conditionalSum(arr, cond) {
  var total_length = arr.length;
  var total_num = 0;
  if (total_length === 0) {
    return 0;
  }
  if (cond === 'even') {
    for (i = 0; i < arr.length; i++) {
      if (arr[i] % 2 === 0) {
        total_num += arr[i];
      }
    }
  }
  if (cond == 'odd') {
    for (i = 0; i < arr.length; i++) {
      if (arr[i] % 2 != 0) {
        total_num += arr[i];
      }
    }
  }

  return total_num;
}

    看一下这个

    回复于
  • -2

    简单而整洁的解决方案

    function conditionalSum(nums, operation) {
  var condition = ['even', 'odd'];
  sum = 0;
  nums.forEach(function(i) {
    if (condition[i % 2] == operation) {
      sum += i;
    }
  });
  return sum;
}

console.log(conditionalSum([1, 2, 3, 4, 5], "even"));
console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));
console.log(conditionalSum([13, 88, 12, 44, 99], "even"));
console.log(conditionalSum([], "odd"));

    希望，这会有所帮助 .

    回复于
  • 0

    我建议将条件函数作为参数传递 .

    这样，您可以为许多不同的操作重用相同的功能 .

    /**
 * conditionalSum
 * Runs a list of numbers against a custom conditional function and accumulators the values that passes
 * @param {number[]} list
 * @param {(n: number) => boolean} condition
 * @returns {number}
 */
function conditionalSum(list, condition) {
    var total = 0;
    for (var index = 0; index < list.length; index++) {
        var value = list[index];
        if (condition(value)) {
            total += value;
        }
    }
    return total;
}
//TEST
function isEven(n) { return n % 2 === 0; }
function isOdd(n) { return n % 2 !== 0; }
console.log(conditionalSum([1, 2, 3, 4, 5], isEven));
console.log(conditionalSum([1, 2, 3, 4, 5], isOdd));
console.log(conditionalSum([13, 88, 12, 44, 99], isEven));
console.log(conditionalSum([], isOdd));
    回复于
  • 0
    const conditionalSum = (numbers, operation) => {
      const value = operation === "even" ? 0 : 1;
      return numbers.filter(number => number % 2 === value).reduce((acc, num) => acc + num, 0);
    }
    
    console.log(conditionalSum([1, 2, 3, 4, 5], "even"));      //   6
    console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));       //   9
    console.log(conditionalSum([13, 88, 12, 44, 99], "even")); // 144
    console.log(conditionalSum([], "odd"));                    //   0
    回复于
  • 0

    实际上有几种方法可以解决这个问题 .

    第一个是使用 filter 获取预期数字列表，然后将它们与 reduce 相加 . 这是为了获得最佳可读性 .

    let conditionalSum = (arr, parity) => {
  return arr
      .filter(num => parity === 'even' ? num % 2 === 0 : num % 2 !== 0)
      .reduce((prev, curr) => prev + curr, 0);
};
console.log(conditionalSum([1, 2, 3, 4, 5], "even"));
console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));
console.log(conditionalSum([13, 88, 12, 44, 99], "even"));
console.log(conditionalSum([], "odd"));

    同时使用 for 循环有助于提供更高性能的解决方案 .

    let conditionalSum = (arr, parity) => {
  let sum = 0;
  for(let num of arr) {
    if(parity === 'even' && num % 2 !== 0
    || parity === 'odd' && num % 2 === 0) {
      continue;
    }
    sum += num;
  }
  return sum;
};

console.log(conditionalSum([1, 2, 3, 4, 5], "even"));
console.log(conditionalSum([1, 2, 3, 4, 5], "odd"));
console.log(conditionalSum([13, 88, 12, 44, 99], "even"));
console.log(conditionalSum([], "odd"));

    因此，选择正确的解决方案取决于您的使用案例 . 我认为大多数时候你应该选择可读性 .

    回复于

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