Question

我有一个要求，用户有自己的基数，并会购买更多的计数产品。

我有折扣系统：

var b1 = { min: 1, max: 5, discount: 0.01 }
var b2 = { min: 6, max: 10, discount: 0.02 }
var b3 = { min: 11, max: 15, discount: 0.03 }
var b4 = { min: 16, max: 20, discount: 0.04 }

var b = [b1, b2, b3, b4]   // this is the discount system

有一个要求，应该是get_discounts方法，通过它我可以得到新的购买产品＆＃39;折扣。

像这样：

var a = 5  // this is the base count for user own.
console.log(get_discount(a, 7, b))  // the 7 is new buy. I want to get the discounts of the 7

// [
//   { count: 2, discount: 0.03 }  // 11, 12 ( 11  <= a + 7 <=15 )
//   { count: 5, discount: 0.02 }  //  ( 6  <= a + 7 <=10 )
// ]
// there exclude the `1 -> 5` (exclude the `a` for discount) . the list counts are 2, 5 (2+5 = 7)

我试过了：

function get_discount(a , new_count, b) {

  var discounts = []

  var tem_count = a + new_count

  for(index in b){

    var item = b[index]

    if (tem_count >= item.min && tem_count <= item.max){

      var discount_obj = {
        count: tem_count - item.min + 1,
        disocunt: item.discount
      }

      tem_count -= discount_obj.count

      return discount_obj
    }
  }

}

我只得到

{ count: 2, disocunt: 0.03 }

有一个明显的问题，而不是预期的效果。

谁可以帮助优化功能？

Answer 1

您可以使用Array#reduceRight。

>>> np.repeat(np.repeat(np.random.randint(0, 256, (4, 4)), 2, 0), 2, 1)
array([[ 41,  41,  29,  29, 103, 103,  67,  67],
       [ 41,  41,  29,  29, 103, 103,  67,  67],
       [231, 231, 203, 203, 231, 231, 157, 157],
       [231, 231, 203, 203, 231, 231, 157, 157],
       [ 18,  18, 126, 126,  15,  15, 196, 196],
       [ 18,  18, 126, 126,  15,  15, 196, 196],
       [198, 198, 152, 152,  74,  74, 211, 211],
       [198, 198, 152, 152,  74,  74, 211, 211]])

function getDiscount(offset, amount, table) {
    var value = offset + amount;
    return table.reduceRight((r, { min, max, discount }) => {
        var count = value - min + 1;
        if (value > amount && min <= value && value <= max) {
            r.push({ count, discount });
            value -= count;
        }
        return r;
    }, []);
}

var discounts = [{ min: 1, max: 5, discount: 0.01 }, { min: 6, max: 10, discount: 0.02 }, { min: 11, max: 15, discount: 0.03 }, { min: 16, max: 20, discount: 0.04 }],
    oldBuy = 5,
    newBuy = 7,
    discount = getDiscount(oldBuy, newBuy, discounts);

console.log(discount);

Answer 2

您可以使用while循环计算新购买次数：

var b1 = { min: 1, max: 5, discount: 0.01 }
var b2 = { min: 6, max: 10, discount: 0.02 }
var b3 = { min: 11, max: 15, discount: 0.03 }
var b4 = { min: 16, max: 20, discount: 0.04 }

var b = [b1, b2, b3, b4]

function get_discount_by_itemcount(tem_count, b) {
  for(index in b){

    var item = b[index]

    if (tem_count >= item.min && tem_count <= item.max){

      var discount_obj = {
        count: tem_count - item.min + 1,
        disocunt: item.discount
      }

      tem_count -= discount_obj.count

      return {
        discount_obj: discount_obj,
        tem_count: tem_count
      }
    }
  }
}

function get_discount(a , new_count, b) {

  var discounts = []

  var tem_count = a + new_count

  while( tem_count > a){

    var obj = get_discount_by_itemcount(tem_count, b)
    tem_count = obj.tem_count
    discounts.push(obj.discount_obj)

  }

  return discounts
}

var a = 5
console.log(get_discount(a, 7, b))

获取级别折扣算法

2 个答案: