Question

由于某种原因，price.index（y）和x给了我奇怪的答案

def stockpicker prices
    profits, buyday, sellday = 0, 0, 0

    i=0
    while i <= prices.length
        y = i + 1
        while y <= prices.length
            if prices[y].to_i - prices[i].to_i > profits
                profits = prices[y].to_i - prices[i].to_i
                sellday = prices.index(y).to_i
                buyday = prices.index(i).to_i
            end
            y += 1
        end
        i += 1
    end

    print "Profits are #{profits} if you buy on #{buyday} and sell on #{sellday}"

end

stockpicker([17,3,6,9,15,8,6,9,1])

=＆gt;如果您在8买入并在0
卖出，则利润为13

为什么不提供正确的索引位置？我真的不知道8和0来自哪里

Answer 1

假设你想要这个输出：

Profits are 12 if you buy on 1 and sell on 4

然后你应该将sellday / buyday设置为当前索引值：

sellday = y
buyday = i

使用sellday = prices.index(y).to_i，您说“给我数组中项目的索引，其中值为4”。这导致nil。 nil.to_i将始终为您提供0。

Answer 2

现在您的问题已经得到解答，让我们看看如何以更加Rubylike的方式编写代码。有很多方法可以做到这一点。这是一个。

<强>代码

def stockpicker(prices)
  a = Array.new(prices.size) { |i|
        Array.new(prices.size) { |j|
          (j > i) ? prices[j]-prices[i] : -Float::INFINITY } }.flatten
  max_profit = a.max
  bd, sd = a.index(max_profit).divmod(prices.size)
  print "Profits are #{max_profit} if you buy on #{bd} and sell on #{sd}"
end

示例

prices = [17,3,6,9,15,8,6,9,1] stockpicker(prices) #=> Profits are 12 if you buy on 1 and sell on 4

<强>解释

对于上面给出的例子：

a = Array.new(prices.size) { |i| Array.new(prices.size) { |j| (j >= i) ? prices[j]-prices[i] : -Float::INFINITY } } #=> [[0, -14, -11, -8, -2, -9, -11, -8, -16], # [-Infinity, 0, 3, 6, 12, 5, 3, 6, -2], # [-Infinity, -Infinity, 0, 3, 9, 2, 0, 3, -5], # [-Infinity, -Infinity, -Infinity, 0, 6, -1, -3, 0, -8], # [-Infinity, -Infinity, -Infinity, -Infinity, 0, -7, -9, -6, -14], # [-Infinity, -Infinity, -Infinity, -Infinity, -Infinity, 0, -2, 1, -7], # [-Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, # 0, 3, -5], # [-Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, # -Infinity, 0, -8], # [-Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, # -Infinity, -Infinity, 0]]

此数组的行i，列j中的条目等于股票在日偏移量i上购买并在当天出售j时的利润。例如，考虑行（偏移）1，它对应于日偏移1上的股票购买：

[-Infinity, 0, 3, 6, 12, 5, 3, 6, -2],

如果在j天购买的股票在抵消日1出售，则j列会获得利润。例如，如果在第4天出售，则利润为12.请注意，当卖出日在买入日之前时，我们指定负无穷大的利润。

接下来，展平这个数组：

b = a.flatten #=> [0, -14, -11, -8, -2, -9, -11, -8, -16, # -Infinity, 0, 3, 6, 12, 5, 3, 6, -2, # -Infinity, -Infinity, 0, 3, 9, 2, 0, 3, -5, # -Infinity, -Infinity, -Infinity, 0, 6, -1, -3, 0, -8, # -Infinity, -Infinity, -Infinity, -Infinity, 0, -7, -9, -6, -14, # -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, 0, -2, 1, -7, # -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, 0, # 3, -5, # -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, # -Infinity, 0, -8, # -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, -Infinity, # -Infinity, -Infinity, 0]

确定最大利润：

max_profit = b.max #=> 12

现在我们需要在展平数组中找到12的索引（如果有多个12，则获取第一个的索引）：

c = b.index(max_profit) #=> b.index(12) => 13

然后使用Fixnum#divmod将此偏移量转换为unflattened数组中的行和列：

bd, sd = c.divmod(prices.size) #=> c.divmod(9) => [1, 4] => bd = 1, sd = 4

Answer 3

以下是使用matrix操作的另一种方式：

def stockpicker(prices)
  m = Matrix.build(prices.size){ |x,y| prices[y] - prices[x]}
  max = m.each(:upper).max
  bd,sd = m.index(max, :upper)
  puts "Profits are #{max} if you buy on #{bd} and sell on #{sd}"
end

示例运行：

prices = [17,3,6,9,15,8,6,9,1]
stockpicker(prices)
#=> Profits are 12 if you buy on 1 and sell on 4

说明：这构建了一个方形矩阵，每个价格从每个价格中减去。由于您只想从较早的日期中减去后几天，因此您只需要矩阵的上三角形（因此:upper）。有内置的Matrix函数来获取最大值及其索引。

array.index（y）没有在Ruby中给出预期的结果

3 个答案: