JavaScript计算找零（freeCodeCamp项目）

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需求

编写一个用于收银机的函数 checkCashRegister()：它的第一个参数为售价 price、第二个参数为支付金额 cash、第三个参数为收银机內的金额 cid。cid 是包含货币面值的二维数组。

函数 checkCashRegister() 应返回含有 status 属性和 change 属性的对象。

如果收银机內的金额少于应找回的零钱数，或者无法返回确切的数目时，返回 {status: "INSUFFICIENT_FUNDS", change: []}。

如果收银机內的金额恰好等于应找回的零钱数，返回 {status: "CLOSED", change: [...]}，其中 change 的属性值就是收银机內的金额。

否则，返回 {status: "OPEN", change: [...]}，其中 change 键值是应找回的零钱数，并将找零的面值由高到低排序。

下面是抽屉里现金数组示例：

[["PENNY", 1.01],
  ["NICKEL", 2.05],
  ["DIME", 3.1],
  ["QUARTER", 4.25],
  ["ONE", 90],
  ["FIVE", 55],
  ["TEN", 20],
  ["TWENTY", 60],
  ["ONE HUNDRED", 100]
]

代码实现

function checkCashRegister(price, cash, cid) {
  let change = cash - price
  let response = {status: '', change: [] }
  // 当前收银机內总金额
  let cidAmount = cid.reduce((accumulator, current) => accumulator + current[1], 0)

  // 收银机內的金额 < 应找回的零钱数
  if (cidAmount < change) {
    response.status = 'INSUFFICIENT_FUNDS'
    return response
  }

  // 收银机內的金额 = 应找回的零钱数
  if (cidAmount === change) {
    response.status = 'CLOSED'
    response.change = cid
    return response
  }

  cid = cid.reverse()
  let moneyTable = []
  let amounts = [100, 20, 10, 5, 1, 0.25, 0.1, 0.05, 0.01]

  for (let i = 0; i < cid.length; i++) {let item = {}
    item['name'] = cid[i][0]
    item['amount'] = amounts[i]
    item['totalAmount'] = cid[i][1]
    moneyTable.push(item)
  }

  for (let i = 0; i < moneyTable.length; i++) {let ele = []

    if (change >= moneyTable[i].amount && change > 0 && moneyTable[i].totalAmount > 0) {ele.push(moneyTable[i].name)
      let money = Math.floor(change / moneyTable[i].amount) * moneyTable[i].amount
      if (money > moneyTable[i].totalAmount) {money = moneyTable[i].totalAmount
      }
      ele.push(money)
      change -= money
      change = +change.toFixed(2)
      response.change.push(ele)
    }

    if (change === 0) {break}
  }

  if (change === 0) {response.status = 'OPEN'} else {
    response.status = 'INSUFFICIENT_FUNDS'
    response.change = []}

  return response
}