Question

我创建了此函数，但最后两个测试

checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 1], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]) should return {status: "INSUFFICIENT_FUNDS", change: []}.
checkCashRegister(19.5, 20, [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]) should return {status: "CLOSED", change: [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]}.

不及格。我的代码有什么问题？

任务是：

设计一个收银机抽屉功能checkCashRegister()，它接受购买价格作为第一个参数（price），付款作为第二个参数（cash）和抽屉现金（ cid）作为第三个参数。

cid是列出可用货币的2D数组。

checkCashRegister()函数应始终返回具有status键和change键的对象。

如果取款现金少于到期的找零，或者您不能退还确切的找零，则退还{status: "INSUFFICIENT_FUNDS", change: []}。

返回{状态：“已关闭”，更改：[...]}，如果它等于到期的更改，则用现金抽屉作为键change的值。

否则，返回{status: "OPEN", change: [...]}，并将应找零的硬币和纸币以change键的值从高到低的顺序排序。

我的代码：

const INSUFFICIENT_FUNDS = {
  status: 'INSUFFICIENT_FUNDS',
  change: []
};
const CLOSED_CID = {
  status: 'CLOSED',
  change: []
};
const POSITION_CID = [
  ['ONE HUNDRED', 100],
  ['TWENTY', 20],
  ['TEN', 10],
  ['FIVE', 5],
  ['ONE', 1],
  ['QUARTER', 0.25],
  ['DIME', 0.1],
  ['NICKEL', 0.05],
  ['PENNY', 0.01]
];
const add = (a, b) => a + b[1];
const checkCashRegister = (price, cash, cid) => {
  let finalChange = [];
  let changeDue = cash - price;
  const cidSum = cid.reduce(add, 0).toFixed(2);
  /*
  If there is exactly enough money to provide change, the
  status key is “CLOSED”, and the change key is our
  cash-in-drawer(cid).
  */
  if (cidSum == changeDue) {
    return {
      CLOSED_CID,
      change: cid
    };
  }
  /*
  If there is not enough money to provide change, the status key
  is “INSUFFICIENT_FUNDS” and the change key is an empty array.
  */
  if (cidSum < changeDue) {
    return INSUFFICIENT_FUNDS;
  }
  /*
  If there is enough money to provide change with money still
  left in the drawer, the change is then provided by going down
  a list of currency units from high to low, pushing them to
  the change array, and subtracting them from both the cash
  owed and the cash-in-drawer(cid).
  */
  let cidReverse = cid.reverse();
  POSITION_CID.map((posCid, index) => {
    const billValue = posCid[1];
    while (changeDue >= billValue && cidReverse[index][1] >= billValue) {
      changeDue -= billValue; // minus change due from bill value
      changeDue = changeDue.toFixed(2); // fix rounding errors
      const hasBillValueAlready = finalChange.filter(p => p[0] === cidReverse[index][0]);
      if (hasBillValueAlready.length > 0) {
        finalChange = finalChange.map(k => (k[0] === posCid[0] && [k[0], (k[1] += billValue)]) || [k[0], k[1]]);
      } else {
        finalChange.push([cidReverse[index][0], billValue]);
      }
      cidReverse[index][1] -= billValue; // minus bill value from cash-in-drawer
    }
  });
  if (changeDue !== 0) {
    return {
      status: 'OPEN',
      change: finalChange
    };
  } else {
    return INSUFFICIENT_FUNDS;
  }
};
const price = 19.5;
const cash = 200;
const cid = [
  ['PENNY', 1.01],
  ['NICKEL', 2.05],
  ['DIME', 3.1],
  ['QUARTER', 4.25],
  ['ONE', 90],
  ['FIVE', 55],
  ['TEN', 20],
  ['TWENTY', 60],
  ['ONE HUNDRED', 100]
];
// checkCashRegister(price, cash, cid);

console.log(
checkCashRegister(19.5, 20, [
  ["PENNY", 1.01],
  ["NICKEL", 2.05],
  ["DIME", 3.1],
  ["QUARTER", 4.25],
  ["ONE", 90],
  ["FIVE", 55],
  ["TEN", 20],
  ["TWENTY", 60],
  ["ONE HUNDRED", 100]
]),
checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 1], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]),"should return", `{status: "INSUFFICIENT_FUNDS", change: []}`,
checkCashRegister(19.5, 20, [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]), "should return", `{status: "CLOSED", change: [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]}`)

Answer 1

要解决的两个问题：

当到期的找零正好在收银机中时，代码将返回一个结构不正确的对象：
```
return {
  CLOSED_CID,
  change: cid
};
```
这将创建属性CLOSED_CID，该属性的值是带有嵌套status和change属性的对象...然后在该属性处添加另一个change属性父级。相反，您应该忘记预定义的CLOSED_CID对象，然后执行以下操作：
```
return {
  status: "CLOSED",
  change: cid
};
```

决定返回状态OPEN还是INSUFFICIENT_FUNDS的条件应该相反。当您将找零金额的 all 全部归因于帐单时，您应该返回OPEN。所以改变这个：

if (changeDue !== 0) {
  return {
    status: 'OPEN',
    change: finalChange
  };
} else {
  return INSUFFICIENT_FUNDS;
}

收件人：

if (changeDue == 0) { // All change could be translated to available bills/coins
  return {
    status: 'OPEN',
    change: finalChange
  };
} else { // There is change remaining that cannot be covered by available bills/coins
  return INSUFFICIENT_FUNDS;
}

请注意：toFixed方法返回一个字符串。最好将结果显式转换回数字数据类型。目前，您的代码中没有问题，因为在执行-=和==时隐式地进行了转换，但是出于代码维护的原因，您最好不要依赖于此。因此，每当您呼叫+时，添加一元toFixed：

const cidSum = +cid.reduce(add, 0).toFixed(2);
// ...
changeDue = +changeDue.toFixed(2);

但是我建议所有的计算都以美分为单位，因此所有的计算都是使用整数值进行的，这样就不会受到不精确性的困扰。

我假设您已经找到了解决方法on the web，并且有many Q&A on Stack Overflow遇到了同样的挑战。

我为Freecodecamp创建了一个“现金注册”功能，但是由于某些原因，最后两个测试未通过？

1 个答案: