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javascript - 两个数组中最接近的值的集合

问题描述

我正在寻找一种算法的优化来解决一个可能难以解释的简单问题。当您阅读代码时，我不是在寻找速度或性能，而是在寻找简单性和清晰度。也许有人有比我更聪明的解决方案。我想单行可能是一种矫枉过正。

我有两个按日期排序的单元格集合。每个单元格都可以有一个价格值。我们可以假设同一日期的两个单元格中不可能有价格。我想要一个日期集合，但是日期没有价格：

这是我到目前为止所拥有的（它给出了准确的结果）：

const array1 = [
  { date: '2019-11-10' },
  { date: '2019-11-11' },
  { date: '2019-11-12' },
  { date: '2019-11-13' },
  { date: '2019-11-14' },
  { date: '2019-11-15', price: 10 },
  { date: '2019-11-16' },
];

const array2 = [
  { date: '2019-11-10' },
  { date: '2019-11-11' },
  { date: '2019-11-12', price: 10 },
  { date: '2019-11-13' },
  { date: '2019-11-14' },
  { date: '2019-11-15' },
  { date: '2019-11-16' },
];

const merged = Object.values(array1).map((element, index) => {
  let filled;
  if (element.price) {
    filled = 1;
  }
  if (array2[index].price) {
    filled = 2;
  }
  if (filled) {
    return {
      date: element.date,
      filled
    }
  } else {
    return {
      date: element.date
    }
  }
});

const first = merged.find(element => element.filled);

let currentFill = first && first.filled;

const emptyMap = merged.map((element, index, array) => {
  if (!element.filled) {
    return {
      date: element.date,
      empty: currentFill
    }
  }
  currentFill = element.filled;
  return element;
})

console.log(emptyMap);

标签： javascriptarraysalgorithmoptimization

解决方案

const array1 = [
  { date: '2019-11-10' },
  { date: '2019-11-11' },
  { date: '2019-11-12' },
  { date: '2019-11-13' },
  { date: '2019-11-14' },
  { date: '2019-11-15', price: 10 },
  { date: '2019-11-16' },
];

const array2 = [
  { date: '2019-11-10' },
  { date: '2019-11-11' },
  { date: '2019-11-12', price: 10 },
  { date: '2019-11-13' },
  { date: '2019-11-14' },
  { date: '2019-11-15' },
  { date: '2019-11-16' },
];

function collect(collection1, collection2) {
  let firstEmptyId = null
  let currentId = null
  let indexes = [/* [1] optimization */]
  let collection = collection1.map(({ date, price }, i) => (
    ((price && (currentId = 1)) || (collection2[i].price && (currentId = 2)))
      ? ((firstEmptyId || (firstEmptyId = currentId)), { date, filled: currentId })
      : ((firstEmptyId || /*see[1]*/ indexes.push(i)), { date, empty: currentId })
  ))
  // only two iterations => index.length === 2
  indexes.forEach((i) => (collection[i].empty = firstEmptyId))
  return collection
}

console.log(collect(array1, array2))

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