Question

此算法的目的是返回更改的有组织数组（包含货币）。

我构建了一个嵌套的while循环来循环，只要＆＃34;更改＆＃34;不等于0然后while循环函数中的每个循环，而＆＃34;更改＆＃34;大于0（即：不是-ve）。

我错过了什么？

// 
function checkCashRegister(price, cash, cid) {
  var change = cash-price;
  var chArr = [["PENNY", 0],
["NICKEL", 0],
["DIME", 0],
["QUARTER", 0],
["ONE", 0],
["FIVE", 0],
["TEN", 0],
["TWENTY", 0],
["ONE HUNDRED", 0]];

  while (change !== 0) {
    while (change - 100 > 0) {
      chArr[8][1] += 100;
      change -=100;
      if (change <= 0) {break;}
    }
    while (change - 20 > 0) {
      chArr[7][1] += 20;
      change -=20;
      if (change <= 0) {break;}
    }
    while (change - 10 > 0) {
      chArr[6][1] += 10;
      change -=10;
      if (change <= 0) {break;}
    }
    while (change - 5 > 0) {
      chArr[5][1] += 5;
      change -=5;
      if (change <= 0) {break;}
    }
    while (change - 1 > 0) {
      chArr[4][1] += 1;
      change -=1;
      if (change <= 0) {break;}
    }
    while (change - 0.25 > 0) {
      chArr[3][1] += 0.25;
      change -=0.25;
      if (change <= 0) {break;}
    }
    while (change - 0.1 > 0) {
      chArr[2][1] += 0.1;
      change -=0.1;
      if (change <= 0) {break;}
    }
    while (change - 0.05 > 0) {
      chArr[1][1] += 0.05;
      change -=0.05;
      if (change <= 0) {break;}
    }
    while (change - 0.01 > 0) {
      chArr[0][1] += 0.01;
      change -=0.01;
      if (change <= 0) {break;}
    }
    if (change <= 0) {break;}

  }
  // Here is your change, ma'am.
  return chArr;
}

checkCashRegister(17.46, 20.00, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]]);

Answer 1

您只需检查单个现金部件，并在那里仅使用检查其余部分是否大于更改。在while循环中不需要破坏条件。

缺少的是检查更改是否足够。

function checkCashRegister(price, cash, cid) {
  var change = cash-price;
  var chArr = [["PENNY", 0], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]];

    while (change >= 100 && cid[8][1] >= 100) {
      chArr[8][1] += 100;
      cid[8][1] -= 100;
      change -=100;
    }
    while (change >= 20 && cid[7][1] >= 20) {
      chArr[7][1] += 20;
      cid[7][1] -= 20;
      change -=20;
    }
    while (change >= 10 && cid[6][1] >= 10) {
      chArr[6][1] += 10;
      cid[6][1] -= 10;
      change -=10;
    }
    while (change >= 5 && cid[5][1] >= 5) {
      chArr[5][1] += 5;
      cid[5][1] -= 5;
      change -=5;
    }
    while (change >= 1 && cid[4][1] >= 1) {
      chArr[4][1] += 1;
      cid[4][1] -= 1;
      change -=1;
    }
    while (change >= 0.25 && cid[3][1] >= 0.25) {
      chArr[3][1] += 0.25;
      cid[3][1] -= 0.25;
      change -=0.25;
    }
    while (change >= 0.1 && cid[2][1] >= 0.1) {
      chArr[2][1] += 0.1;
      cid[2][1] -= 0.1;
      change -=0.1;
    }
    while (change >= 0.05 && cid[1][1] >= 0.05) {
      chArr[1][1] += 0.05;
      cid[1][1] -= 0.05;
      change -=0.05;
    }
    while (change > 0 && cid[0][1] >= 0.01) {
      chArr[0][1] += 0.01;
      cid[0][1] -= 0.01;
      change -=0.01;
    }
  
  // Here is your change, ma'am.
  return chArr;
}

var register1 = [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]],
    register2 = [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]];

console.log(checkCashRegister(17.46, 20.00, register1));
console.log(register1);

console.log(checkCashRegister(19.50, 20.00, register2));
console.log(register2);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

我认为您应该重新考虑您的代码：

function checkCashRegister(price, cash) {
  var change = cash - price;

  var chObj = {
    PENNY: 0,
    NICKEL: 0,
    QUARTER: 0,
    ONE: 0,
    FIVE: 0,
    TEN: 0,
    TWENTY: 0,
    ONE_HUNDRED: 0
  }

  // Performs an euclidian division to get the quotient
  // (in 750$, there are 7 times 100 ie quotient = 7)    
  chObj.ONE_HUNDRED = Math.floor(change / 100);
  // Now get the remainder (750 minus 7 times 100)
  change = change % 100;

  // So on...
  chObj.TWENTY = Math.floor(change / 20);
  change = change % 20;

  chObj.TEN = Math.floor(change / 10);
  change = change % 10;

  chObj.FIVE = Math.floor(change / 5);
  change = change % 5;

  chObj.ONE = Math.floor(change);

  change = Math.floor((change - Math.floor(change)) * 100);

  chObj.QUARTER = Math.floor(change / 20) / 100;
  change = change % 25;

  chObj.NICKEL = Math.floor(change / 10) / 100;
  change = change % 10;

  chObj.NICKEL = Math.floor(change) / 100;

  return chObj;
}

var chObj = checkCashRegister(245, 1000);

for (var key in chObj) {
  if (chObj.hasOwnProperty(key)) {
    console.log(key + " : \t" + chObj[key]);
  }
}

结果（可通过控制台查看）：

PENNY :     0
NICKEL :    0
QUARTER :   0
ONE :   0
FIVE :  1
TEN :   1
TWENTY :    2
ONE_HUNDRED :   7

我认为可以进一步改善。请参阅小提琴here。

Answer 3

您的问题是，您正在检查等式为0，而您的实际值为0.00 .. [某事]您可以看到它是一个已知问题here

将最后一个if语句更改为if (change.toFixed(2) <= 0) {break;}

BTW我建议你把你所有的钱计算都改为美分，以避免这种问题（而不是1.0美元使用100美分）。

Answer 4

原因如下：每种类型的最终硬币都不会添加到阵列中。

说金额是0.02,2美分。然后在最后一个循环中，

while (change - 0.01 > 0) {
  chArr[0][1] += 0.01;
  change -=0.01;
  if (change <= 0) {break;}
}

守卫（0.02 - 0.01 > 0）为真，因此输入循环并且变为0.01。

下次，0.01 - 0.01 > 0 false ，因此未输入循环。

然后外循环一次又一次地运行，因为变化仍然是0.01并且它永远不会变低。

因此>应该是>=。然后可以删除所有断点以及整个外循环。

无限循环虽然我休息了;

4 个答案: