最大单笔销售利润
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【中文标题】最大单笔销售利润【英文标题】：Maximum single-sell profit 【发布时间】：2011-10-28 12:36:14 【问题描述】：假设我们有一个 n 个整数数组，表示一天的股票价格。我们希望找到一对 (buyDay, sellDay)，其中 buyDay ≤ sellDay，这样如果我们在 buyDay 买入股票并卖出在sellDay，我们将最大化我们的利润。
通过尝试所有可能的 (buyDay, sellDay) 对并采用最好的。但是，有没有更好的算法，或许可以在 O(n) 时间内运行？
【问题讨论】：
这是一个间接级别的最大和子序列问题。 @MSN：怎么会这样？他根本不看总和，而是看元素之间的差异。 @PengOne，就像我说的，它有一层间接性。具体来说，您希望在连续几天内最大化收益/损失的总和。因此，将列表转换为收益/损失，并找到最大的子序列和。 为什么你不能找到最小值和最大值并返回呢？我对这个问题有什么误解？ @PDN ：这是行不通的，因为 min 可能发生在 max 之前。你不能卖股票（在这种情况下），然后再买。 【参考方案1】：我喜欢这个问题。这是一个经典的面试问题，根据你的想法，你最终会得到越来越好的解决方案。当然可以在比 O(n2) 时间更好的时间内完成此操作，我列出了三种不同的方式供您在这里思考问题。希望这能回答您的问题！
首先，分而治之的解决方案。让我们看看我们是否可以通过将输入分成两半来解决这个问题，解决每个子数组中的问题，然后将两者组合在一起。事实证明，我们实际上可以做到这一点，并且可以有效地做到这一点！直觉如下。如果我们只有一天，最好的选择是在当天买入，然后在当天卖回，不盈利。否则，将数组分成两半。如果我们考虑最佳答案可能是什么，它必须位于以下三个位置之一：
-
正确的买入/卖出对完全发生在上半年。
正确的买入/卖出对完全发生在下半年。
正确的买入/卖出对出现在两半 - 我们在上半年买入，然后在下半年卖出。
我们可以通过在前半部分和后半部分递归调用我们的算法来获得 (1) 和 (2) 的值。对于方案（3），获得最高利润的方法是在上半年的最低点买入，在下半年的最高点卖出。我们可以通过对输入进行简单的线性扫描并找到两个值来找到两半中的最小值和最大值。这给了我们一个具有以下递归的算法：
T(1) <= O(1)
T(n) <= 2T(n / 2) + O(n)
使用Master Theorem 解决递归问题，我们发现这在 O(n lg n) 时间内运行，并且将使用 O(lg n) 空间进行递归调用。我们刚刚击败了简单的 O(n2) 解决方案！
但是等等！我们可以做得比这更好。请注意，我们在递归中使用 O(n) 项的唯一原因是我们必须扫描整个输入以试图找到每一半的最小值和最大值。由于我们已经在递归地探索每一半，也许我们可以通过让递归也返回存储在每一半中的最小值和最大值来做得更好！换句话说，我们的递归返回了三件事：
-
实现利润最大化的买卖时间。
范围内的整体最小值。
范围内的最大值。
这最后两个值可以使用简单的递归进行递归计算，我们可以在递归计算 (1) 的同时运行：
-
单个元素范围的最大值和最小值就是那个元素。
可以通过将输入分成两半，找到每一半的最大值和最小值，然后取它们各自的最大值和最小值来找到多元素范围的最大值和最小值。
如果我们使用这种方法，我们的递归关系现在是
T(1) <= O(1)
T(n) <= 2T(n / 2) + O(1)
在这里使用主定理给了我们 O(n) 的运行时间和 O(lg n) 的空间，这甚至比我们原来的解决方案更好！
但是等一下 - 我们可以做得比这更好！让我们考虑使用动态规划来解决这个问题。我们的想法是如下思考问题。假设我们在查看了前 k 个元素后就知道了问题的答案。我们能否利用我们对第 (k+1) 个元素的知识，结合我们的初始解决方案，来解决第一个 (k+1) 个元素的问题？如果是这样，我们可以通过解决第一个元素的问题，然后是前两个，然后是前三个等，直到我们为前 n 个元素计算它，从而得到一个很好的算法。
让我们考虑一下如何做到这一点。如果我们只有一个元素，我们已经知道它必须是最好的买卖对。现在假设我们知道前 k 个元素的最佳答案并查看第 (k+1) 个元素。那么，这个值可以创建比我们对前 k 个元素更好的解决方案的唯一方法是，如果前 k 个元素中的最小元素与新元素之间的差异大于我们迄今为止计算的最大差异。所以假设当我们遍历元素时，我们跟踪两个值——到目前为止我们看到的最小值，以及仅使用前 k 个元素可以获得的最大利润。最初，我们到目前为止看到的最小值是第一个元素，最大利润为零。当我们看到一个新元素时，我们首先通过计算以目前看到的最低价格买入并以当前价格卖出来获得多少来更新我们的最佳利润。如果这比我们迄今为止计算的最优值要好，那么我们将最优解更新为这个新的利润。接下来，我们将目前看到的最小元素更新为当前最小元素和新元素中的最小值。
因为在每个步骤中我们只做 O(1) 的工作并且我们只访问了 n 个元素中的每一个元素一次，所以这需要 O(n) 的时间来完成！此外，它仅使用 O(1) 辅助存储。到目前为止，这和我们所做的一样好！
例如，在您的输入中，该算法的运行方式如下。数组中每个值之间的数字对应于算法在该点保存的值。您实际上不会存储所有这些（这将占用 O(n) 内存！），但看到算法的发展会很有帮助：
5 10 4 6 7
min 5 5 4 4 4
best (5,5) (5,10) (5,10) (5,10) (5,10)
答案：(5, 10)
5 10 4 6 12
min 5 5 4 4 4
best (5,5) (5,10) (5,10) (5,10) (4,12)
答案：(4, 12)
1 2 3 4 5
min 1 1 1 1 1
best (1,1) (1,2) (1,3) (1,4) (1,5)
答案：(1, 5)
我们现在可以做得更好吗？不幸的是，不是渐近意义上的。如果我们使用少于 O(n) 的时间，我们就无法查看大型输入上的所有数字，因此无法保证不会错过最佳答案（我们可以将其“隐藏”在我们的元素中）没看）。另外，我们不能使用少于 O(1) 的空间。可能对隐藏在 big-O 符号中的常数因子进行了一些优化，但除此之外，我们无法期望找到任何更好的选择。
总的来说，这意味着我们有以下算法：
朴素：O(n2) 时间，O(1) 空间。 分而治之：O(n lg n) 时间，O(lg n) 空间。 优化的分而治之：O(n) 时间，O(lg n) 空间。 动态编程：O(n) 时间，O(1) 空间。希望这会有所帮助！
编辑：如果您有兴趣，我已经编写了 a Python version of these four algorithms 代码，以便您可以与他们一起玩耍并判断他们的相对表现。代码如下：
# Four different algorithms for solving the maximum single-sell profit problem,
# each of which have different time and space complexity. This is one of my
# all-time favorite algorithms questions, since there are so many different
# answers that you can arrive at by thinking about the problem in slightly
# different ways.
#
# The maximum single-sell profit problem is defined as follows. You are given
# an array of stock prices representing the value of some stock over time.
# Assuming that you are allowed to buy the stock exactly once and sell the
# stock exactly once, what is the maximum profit you can make? For example,
# given the prices
#
# 2, 7, 1, 8, 2, 8, 4, 5, 9, 0, 4, 5
#
# The maximum profit you can make is 8, by buying when the stock price is 1 and
# selling when the stock price is 9. Note that while the greatest difference
# in the array is 9 (by subtracting 9 - 0), we cannot actually make a profit of
# 9 here because the stock price of 0 comes after the stock price of 9 (though
# if we wanted to lose a lot of money, buying high and selling low would be a
# great idea!)
#
# In the event that there's no profit to be made at all, we can always buy and
# sell on the same date. For example, given these prices (which might
# represent a buggy-whip manufacturer:)
#
# 9, 8, 7, 6, 5, 4, 3, 2, 1, 0
#
# The best profit we can make is 0 by buying and selling on the same day.
#
# Let's begin by writing the simplest and easiest algorithm we know of that
# can solve this problem - brute force. We will just consider all O(n^2) pairs
# of values, and then pick the one with the highest net profit. There are
# exactly n + (n - 1) + (n - 2) + ... + 1 = n(n + 1)/2 different pairs to pick
# from, so this algorithm will grow quadratically in the worst-case. However,
# it uses only O(1) memory, which is a somewhat attractive feature. Plus, if
# our first intuition for the problem gives a quadratic solution, we can be
# satisfied that if we don't come up with anything else, we can always have a
# polynomial-time solution.
def BruteForceSingleSellProfit(arr):
# Store the best possible profit we can make; initially this is 0.
bestProfit = 0;
# Iterate across all pairs and find the best out of all of them. As a
# minor optimization, we don't consider any pair consisting of a single
# element twice, since we already know that we get profit 0 from this.
for i in range(0, len(arr)):
for j in range (i + 1, len(arr)):
bestProfit = max(bestProfit, arr[j] - arr[i])
return bestProfit
# This solution is extremely inelegant, and it seems like there just *has* to
# be a better solution. In fact, there are many better solutions, and we'll
# see three of them.
#
# The first insight comes if we try to solve this problem by using a divide-
# and-conquer strategy. Let's consider what happens if we split the array into
# two (roughly equal) halves. If we do so, then there are three possible
# options about where the best buy and sell times are:
#
# 1. We should buy and sell purely in the left half of the array.
# 2. We should buy and sell purely in the right half of the array.
# 3. We should buy in the left half of the array and sell in the right half of
# the array.
#
# (Note that we don't need to consider selling in the left half of the array
# and buying in the right half of the array, since the buy time must always
# come before the sell time)
#
# If we want to solve this problem recursively, then we can get values for (1)
# and (2) by recursively invoking the algorithm on the left and right
# subarrays. But what about (3)? Well, if we want to maximize our profit, we
# should be buying at the lowest possible cost in the left half of the array
# and selling at the highest possible cost in the right half of the array.
# This gives a very elegant algorithm for solving this problem:
#
# If the array has size 0 or size 1, the maximum profit is 0.
# Otherwise:
# Split the array in half.
# Compute the maximum single-sell profit in the left array, call it L.
# Compute the maximum single-sell profit in the right array, call it R.
# Find the minimum of the first half of the array, call it Min
# Find the maximum of the second half of the array, call it Max
# Return the maximum of L, R, and Max - Min.
#
# Let's consider the time and space complexity of this algorithm. Our base
# case takes O(1) time, and in our recursive step we make two recursive calls,
# one on each half of the array, and then does O(n) work to scan the array
# elements to find the minimum and maximum values. This gives the recurrence
#
# T(1) = O(1)
# T(n) = 2T(n / 2) + O(n)
#
# Using the Master Theorem, this recurrence solves to O(n log n), which is
# asymptotically faster than our original approach! However, we do pay a
# (slight) cost in memory usage. Because we need to maintain space for all of
# the stack frames we use. Since on each recursive call we cut the array size
# in half, the maximum number of recursive calls we can make is O(log n), so
# this algorithm uses O(n log n) time and O(log n) memory.
def DivideAndConquerSingleSellProfit(arr):
# Base case: If the array has zero or one elements in it, the maximum
# profit is 0.
if len(arr) <= 1:
return 0;
# Cut the array into two roughly equal pieces.
left = arr[ : len(arr) / 2]
right = arr[len(arr) / 2 : ]
# Find the values for buying and selling purely in the left or purely in
# the right.
leftBest = DivideAndConquerSingleSellProfit(left)
rightBest = DivideAndConquerSingleSellProfit(right)
# Compute the best profit for buying in the left and selling in the right.
crossBest = max(right) - min(left)
# Return the best of the three
return max(leftBest, rightBest, crossBest)
# While the above algorithm for computing the maximum single-sell profit is
# better timewise than what we started with (O(n log n) versus O(n^2)), we can
# still improve the time performance. In particular, recall our recurrence
# relation:
#
# T(1) = O(1)
# T(n) = 2T(n / 2) + O(n)
#
# Here, the O(n) term in the T(n) case comes from the work being done to find
# the maximum and minimum values in the right and left halves of the array,
# respectively. If we could find these values faster than what we're doing
# right now, we could potentially decrease the function's runtime.
#
# The key observation here is that we can compute the minimum and maximum
# values of an array using a divide-and-conquer approach. Specifically:
#
# If the array has just one element, it is the minimum and maximum value.
# Otherwise:
# Split the array in half.
# Find the minimum and maximum values from the left and right halves.
# Return the minimum and maximum of these two values.
#
# Notice that our base case does only O(1) work, and our recursive case manages
# to do only O(1) work in addition to the recursive calls. This gives us the
# recurrence relation
#
# T(1) = O(1)
# T(n) = 2T(n / 2) + O(1)
#
# Using the Master Theorem, this solves to O(n).
#
# How can we make use of this result? Well, in our current divide-and-conquer
# solution, we split the array in half anyway to find the maximum profit we
# could make in the left and right subarrays. Could we have those recursive
# calls also hand back the maximum and minimum values of the respective arrays?
# If so, we could rewrite our solution as follows:
#
# If the array has size 1, the maximum profit is zero and the maximum and
# minimum values are the single array element.
# Otherwise:
# Split the array in half.
# Compute the maximum single-sell profit in the left array, call it L.
# Compute the maximum single-sell profit in the right array, call it R.
# Let Min be the minimum value in the left array, which we got from our
# first recursive call.
# Let Max be the maximum value in the right array, which we got from our
# second recursive call.
# Return the maximum of L, R, and Max - Min for the maximum single-sell
# profit, and the appropriate maximum and minimum values found from
# the recursive calls.
#
# The correctness proof for this algorithm works just as it did before, but now
# we never actually do a scan of the array at each step. In fact, we do only
# O(1) work at each level. This gives a new recurrence
#
# T(1) = O(1)
# T(n) = 2T(n / 2) + O(1)
#
# Which solves to O(n). We're now using O(n) time and O(log n) memory, which
# is asymptotically faster than before!
#
# The code for this is given below:
def OptimizedDivideAndConquerSingleSellProfit(arr):
# If the array is empty, the maximum profit is zero.
if len(arr) == 0:
return 0
# This recursive helper function implements the above recurrence. It
# returns a triple of (max profit, min array value, max array value). For
# efficiency reasons, we always reuse the array and specify the bounds as
# [lhs, rhs]
def Recursion(arr, lhs, rhs):
# If the array has just one element, we return that the profit is zero
# but the minimum and maximum values are just that array value.
if lhs == rhs:
return (0, arr[lhs], arr[rhs])
# Recursively compute the values for the first and latter half of the
# array. To do this, we need to split the array in half. The line
# below accomplishes this in a way that, if ported to other languages,
# cannot result in an integer overflow.
mid = lhs + (rhs - lhs) / 2
# Perform the recursion.
( leftProfit, leftMin, leftMax) = Recursion(arr, lhs, mid)
(rightProfit, rightMin, rightMax) = Recursion(arr, mid + 1, rhs)
# Our result is the maximum possible profit, the minimum of the two
# minima we've found (since the minimum of these two values gives the
# minimum of the overall array), and the maximum of the two maxima.
maxProfit = max(leftProfit, rightProfit, rightMax - leftMin)
return (maxProfit, min(leftMin, rightMin), max(leftMax, rightMax))
# Using our recursive helper function, compute the resulting value.
profit, _, _ = Recursion(arr, 0, len(arr) - 1)
return profit
# At this point we've traded our O(n^2)-time, O(1)-space solution for an O(n)-
# time, O(log n) space solution. But can we do better than this?
#
# To find a better algorithm, we'll need to switch our line of reasoning.
# Rather than using divide-and-conquer, let's see what happens if we use
# dynamic programming. In particular, let's think about the following problem.
# If we knew the maximum single-sell profit that we could get in just the first
# k array elements, could we use this information to determine what the
# maximum single-sell profit would be in the first k + 1 array elements? If we
# could do this, we could use the following algorithm:
#
# Find the maximum single-sell profit to be made in the first 1 elements.
# For i = 2 to n:
# Compute the maximum single-sell profit using the first i elements.
#
# How might we do this? One intuition is as follows. Suppose that we know the
# maximum single-sell profit of the first k elements. If we look at k + 1
# elements, then either the maximum profit we could make by buying and selling
# within the first k elements (in which case nothing changes), or we're
# supposed to sell at the (k + 1)st price. If we wanted to sell at this price
# for a maximum profit, then we would want to do so by buying at the lowest of
# the first k + 1 prices, then selling at the (k + 1)st price.
#
# To accomplish this, suppose that we keep track of the minimum value in the
# first k elements, along with the maximum profit we could make in the first
# k elements. Upon seeing the (k + 1)st element, we update what the current
# minimum value is, then update what the maximum profit we can make is by
# seeing whether the difference between the (k + 1)st element and the new
# minimum value is. Note that it doesn't matter what order we do this in; if
# the (k + 1)st element is the smallest element so far, there's no possible way
# that we could increase our profit by selling at that point.
#
# To finish up this algorithm, we should note that given just the first price,
# the maximum possible profit is 0.
#
# This gives the following simple and elegant algorithm for the maximum single-
# sell profit problem:
#
# Let profit = 0.
# Let min = arr[0]
# For k = 1 to length(arr):
# If arr[k] < min, set min = arr[k]
# If profit < arr[k] - min, set profit = arr[k] - min
#
# This is short, sweet, and uses only O(n) time and O(1) memory. The beauty of
# this solution is that we are quite naturally led there by thinking about how
# to update our answer to the problem in response to seeing some new element.
# In fact, we could consider implementing this algorithm as a streaming
# algorithm, where at each point in time we maintain the maximum possible
# profit and then update our answer every time new data becomes available.
#
# The final version of this algorithm is shown here:
def DynamicProgrammingSingleSellProfit(arr):
# If the array is empty, we cannot make a profit.
if len(arr) == 0:
return 0
# Otherwise, keep track of the best possible profit and the lowest value
# seen so far.
profit = 0
cheapest = arr[0]
# Iterate across the array, updating our answer as we go according to the
# above pseudocode.
for i in range(1, len(arr)):
# Update the minimum value to be the lower of the existing minimum and
# the new minimum.
cheapest = min(cheapest, arr[i])
# Update the maximum profit to be the larger of the old profit and the
# profit made by buying at the lowest value and selling at the current
# price.
profit = max(profit, arr[i] - cheapest)
return profit
# To summarize our algorithms, we have seen
#
# Naive: O(n ^ 2) time, O(1) space
# Divide-and-conquer: O(n log n) time, O(log n) space
# Optimized divide-and-conquer: O(n) time, O(log n) space
# Dynamic programming: O(n) time, O(1) space
【讨论】：
@FrankQ.- 两个递归调用都需要空间，但通常这些调用一个接一个地执行。这意味着编译器可以在调用之间重用内存；一旦一个调用返回，下一个调用就可以重用它的空间。因此，您一次只需要内存来保存一个函数调用，因此内存使用量与调用堆栈的最大深度成正比。由于递归在 O(log n) 级别终止，因此只需要使用 O(log n) 内存。这能说明问题吗？ 谁能把这些移植到 Ruby 上？某些递归的工作方式与 Python 中的不同。这些解决方案也只返回最大的利润；他们不返回产生利润的数组点（可用于报告过去利润增长的百分比） 动态规划的概念对于解释 O(n) 时间解决方案并不是真正需要的，但是您可以将所有这些类型的算法结合起来。 如何在任何 sub O(n^2) 算法的基础上找到所有按利润排序的货币对？ @templatetypedef 如果我们从 M$ 的预算开始，而不是单只股票，我们将有 m 只股票，价格超过 n 天，我们将如何改变动态规划方法？即我们改变购买的股票单位数量和可用的股票数据从 1 股到 n 股（就像以前，我们只有谷歌，现在我们也有其他 5 家公司）【参考方案2】：这是带有一点间接性的最大和子序列问题。最大和子序列问题给定一个整数列表，可以是正数或负数，找到该列表的连续子集的最大和。
您可以通过在连续几天之间获取盈亏来轻松地将这个问题转换为那个问题。因此，您将转换股票价格列表，例如[5, 6, 7, 4, 2] 到收益/损失列表中，例如，[1, 1, -3, -2]。子序列和问题就很容易解决了：Find the subsequence with largest sum of elements in an array
【讨论】：
我不认为它完全这样运作，因为如果您在最初的一天购买股票，您不会从前一天累积增量收益日。或者这不是这种方法的问题？ @templatetypedef，这就是你跟踪最大和和当前序列和的原因。当当前序列总和低于零时，您知道您不会通过该序列赚到任何钱，您可以重新开始。通过跟踪最大金额，您将自动找到最佳买入/卖出日期。 @templatetypedef，顺便说一句，你在回答中做了同样的事情。【参考方案3】：我不太确定为什么这被认为是一个动态编程问题。我在教科书和算法指南中看到了这个问题，使用 O(n log n) 运行时和 O(log n) 空间（例如编程面试元素）。这似乎是一个比人们想象的要简单得多的问题。
这通过跟踪最大利润、最低买入价以及最优买入/卖出价来实现。当它遍历数组中的每个元素时，它会检查给定元素是否小于最低购买价格。如果是，则将最低购买价格指数 (min) 更新为该元素的指数。此外，对于每个元素，becomeABillionaire 算法会检查 arr[i] - arr[min]（当前元素与最低买入价之间的差值）是否大于当前利润。如果是，则将利润更新为该差值并将买入设置为arr[min]，将卖出设置为arr[i]。
一次性运行。
static void becomeABillionaire(int arr[])
int i = 0, buy = 0, sell = 0, min = 0, profit = 0;
for (i = 0; i < arr.length; i++)
if (arr[i] < arr[min])
min = i;
else if (arr[i] - arr[min] > profit)
buy = min;
sell = i;
profit = arr[i] - arr[min];
System.out.println("We will buy at : " + arr[buy] + " sell at " + arr[sell] +
" and become billionaires worth " + profit );
合著者：https://***.com/users/599402/ephraim
【讨论】：
【参考方案4】：问题与最大子序列相同 我使用动态编程解决了它。跟踪当前和以前的（利润、买入日期和卖出日期） 如果当前高于先前，则将先前替换为当前。
int prices[] = 38, 37, 35, 31, 20, 24, 35, 21, 24, 21, 23, 20, 23, 25, 27 ;
int buyDate = 0, tempbuyDate = 0;
int sellDate = 0, tempsellDate = 0;
int profit = 0, tempProfit =0;
int i ,x = prices.length;
int previousDayPrice = prices[0], currentDayprice=0;
for(i=1 ; i<x; i++ )
currentDayprice = prices[i];
if(currentDayprice > previousDayPrice ) // price went up
tempProfit = tempProfit + currentDayprice - previousDayPrice;
tempsellDate = i;
else // price went down
if(tempProfit>profit) // check if the current Profit is higher than previous profit....
profit = tempProfit;
sellDate = tempsellDate;
buyDate = tempbuyDate;
// re-intialized buy&sell date, profit....
tempsellDate = i;
tempbuyDate = i;
tempProfit =0;
previousDayPrice = currentDayprice;
// if the profit is highest till the last date....
if(tempProfit>profit)
System.out.println("buydate " + tempbuyDate + " selldate " + tempsellDate + " profit " + tempProfit );
else
System.out.println("buydate " + buyDate + " selldate " + sellDate + " profit " + profit );
【讨论】：
【参考方案5】：这是我的 Java 解决方案：
public static void main(String[] args)
int A[] = 5,10,4,6,12;
int min = A[0]; // Lets assume first element is minimum
int maxProfit = 0; // 0 profit, if we buy & sell on same day.
int profit = 0;
int minIndex = 0; // Index of buy date
int maxIndex = 0; // Index of sell date
//Run the loop from next element
for (int i = 1; i < A.length; i++)
//Keep track of minimum buy price & index
if (A[i] < min)
min = A[i];
minIndex = i;
profit = A[i] - min;
//If new profit is more than previous profit, keep it and update the max index
if (profit > maxProfit)
maxProfit = profit;
maxIndex = i;
System.out.println("maxProfit is "+maxProfit);
System.out.println("minIndex is "+minIndex);
System.out.println("maxIndex is "+maxIndex);
【讨论】：
@Nitiraj ，是的，这个解决方案是正确的，但我要求您阅读 templatetypedef 提供的答案，正如 templatetypedef 提供的答案一样，提到了所有可能的解决方案，包括 Rohit 发布的解决方案. Rohit 的解决方案实际上是使用 templatetypedef 提供的答案中提到的动态编程使用 O(n) 实现最后一个解决方案。 假设你的数组是 int A[] = 5, 4, 6 ,7 ,6 ,3 ,2, 5;然后根据您的逻辑，您将以指数 6 买入，然后以指数 3 卖出。这是错误的。你不能卖过去。卖出指数必须大于买入指数。 上述解决方案“几乎”正确。但它会打印绝对最低指数而不是“买入”价格的指数。要更正，您需要另一个变量，例如 minBuyIndex，您只在“if (profit > maxProfit)”块内更新它并打印它。【参考方案6】：我想出了一个简单的解决方案 - 代码更不言自明。这是那些动态规划问题之一。
代码不处理错误检查和边缘情况。它只是一个示例，给出解决问题的基本逻辑的想法。
namespace MaxProfitForSharePrice
class MaxProfitForSharePrice
private static int findMax(int a, int b)
return a > b ? a : b;
private static void GetMaxProfit(int[] sharePrices)
int minSharePrice = sharePrices[0], maxSharePrice = 0, MaxProft = 0;
int shareBuyValue = sharePrices[0], shareSellValue = sharePrices[0];
for (int i = 0; i < sharePrices.Length; i++)
if (sharePrices[i] < minSharePrice )
minSharePrice = sharePrices[i];
// if we update the min value of share, we need to reset the Max value as
// we can only do this transaction in-sequence. We need to buy first and then only we can sell.
maxSharePrice = 0;
else
maxSharePrice = sharePrices[i];
// We are checking if max and min share value of stock are going to
// give us better profit compare to the previously stored one, then store those share values.
if (MaxProft < (maxSharePrice - minSharePrice))
shareBuyValue = minSharePrice;
shareSellValue = maxSharePrice;
MaxProft = findMax(MaxProft, maxSharePrice - minSharePrice);
Console.WriteLine("Buy stock at $0 and sell at $1, maximum profit can be earned $2.", shareBuyValue, shareSellValue, MaxProft);
static void Main(string[] args)
int[] sampleArray = new int[] 1, 3, 4, 1, 1, 2, 11 ;
GetMaxProfit(sampleArray);
Console.ReadLine();
【讨论】：
【参考方案7】：public static double maxProfit(double [] stockPrices)
double initIndex = 0, finalIndex = 0;
double tempProfit = list[1] - list[0];
double maxSum = tempProfit;
double maxEndPoint = tempProfit;
for(int i = 1 ;i<list.length;i++)
tempProfit = list[ i ] - list[i - 1];;
if(maxEndPoint < 0)
maxEndPoint = tempProfit;
initIndex = i;
else
maxEndPoint += tempProfit;
if(maxSum <= maxEndPoint)
maxSum = maxEndPoint ;
finalIndex = i;
System.out.println(initIndex + " " + finalIndex);
return maxSum;
这是我的解决方案。修改最大子序列算法。解决了 O(n) 中的问题。我认为它不能做得更快。
【讨论】：
【参考方案8】：这是一个有趣的问题，因为它似乎很难，但仔细考虑会产生一个优雅、精简的解决方案。
如前所述，它可以在 O(N^2) 时间内通过暴力破解。对于数组（或列表）中的每个条目，迭代所有先前的条目以获得最小值或最大值，具体取决于问题是找到最大的收益还是损失。
以下是在 O(N) 中考虑解决方案的方法：每个条目代表一个新的可能最大值（或最小值）。然后，我们需要做的就是保存先前的最小值（或最大值），并将差异与当前和先前的最小值（或最大值）进行比较。轻松愉快。
这是代码，在 Java 中作为 JUnit 测试：
import org.junit.Test;
public class MaxDiffOverSeriesProblem
@Test
public void test1()
int[] testArr = new int[]100, 80, 70, 65, 95, 120, 150, 75, 95, 100, 110, 120, 90, 80, 85, 90;
System.out.println("maxLoss: " + calculateMaxLossOverSeries(testArr) + ", maxGain: " + calculateMaxGainOverSeries(testArr));
private int calculateMaxLossOverSeries(int[] arr)
int maxLoss = 0;
int idxMax = 0;
for (int i = 0; i < arr.length; i++)
if (arr[i] > arr[idxMax])
idxMax = i;
if (arr[idxMax] - arr[i] > maxLoss)
maxLoss = arr[idxMax] - arr[i];
return maxLoss;
private int calculateMaxGainOverSeries(int[] arr)
int maxGain = 0;
int idxMin = 0;
for (int i = 0; i < arr.length; i++)
if (arr[i] < arr[idxMin])
idxMin = i;
if (arr[i] - arr[idxMin] > maxGain)
maxGain = arr[i] - arr[idxMin];
return maxGain;
在计算最大损失的情况下，我们会跟踪列表中的最大值（买入价）直到当前条目。然后我们计算最大值和当前条目之间的差异。如果 max - current > maxLoss，那么我们将这个 diff 保留为新的 maxLoss。由于 max 的索引保证小于 current 的索引，我们保证 'buy' 日期小于 'sell' 日期。
在计算最大增益的情况下，一切都颠倒过来。我们跟踪列表中的最小值，直到当前条目。我们计算最小值和当前条目之间的差异（颠倒减法中的顺序）。如果 current - min > maxGain，那么我们将这个 diff 保留为新的 maxGain。同样，“买入”（最小值）的索引位于当前（“卖出”）的索引之前。
我们只需要跟踪 maxGain（或 maxLoss），以及 min 或 max 的索引，但不能同时跟踪两者，并且我们不需要比较索引来验证 'buy' 小于 'sell' ，因为我们自然而然地得到了这个。
【讨论】：
【参考方案9】：最大单笔销售利润，O(n) 解
function stocks_n(price_list)
var maxDif=0, min=price_list[0]
for (var i in price_list)
p = price_list[i];
if (p<min)
min=p
else if (p-min>maxDif)
maxDif=p-min;
return maxDif
这是一个项目，它在 100k 个整数上的随机数据集上对 o(N) 与 o(n^2) 方法进行时间复杂度测试。 O(n^2) 耗时 2 秒，而 O(n) 耗时 0.01s
https://github.com/gulakov/complexity.js
function stocks_n2(ps)
for (maxDif=0,i=_i=0;p=ps[i++];i=_i++)
for (;p2=ps[i++];)
if (p2-p>maxDif)
maxDif=p2-p
return maxDif
这是较慢的 o(n^2) 方法，每天在剩余的日子里循环，双循环。
【讨论】：
【参考方案10】：最高票数的答案不允许最大利润为负数的情况，应进行修改以允许此类情况。可以通过将循环的范围限制为 (len(a) - 1) 并通过将索引移动 1 来改变利润的确定方式来做到这一点。
def singSellProfit(a):
profit = -max(a)
low = a[0]
for i in range(len(a) - 1):
low = min(low, a[i])
profit = max(profit, a[i + 1] - low)
return profit
比较这个版本的函数和之前的数组：
s = [19,11,10,8,5,2]
singSellProfit(s)
-1
DynamicProgrammingSingleSellProfit(s)
0
【讨论】：
【参考方案11】：static void findmaxprofit(int[] stockvalues)
int buy=0,sell=0,buyingpoint=0,sellingpoint=0,profit=0,currentprofit=0;
int finalbuy=0,finalsell=0;
if(stockvalues.length!=0)
buy=stockvalues[0];
for(int i=1;i<stockvalues.length;i++)
if(stockvalues[i]<buy&&i!=stockvalues.length-1)
buy=stockvalues[i];
buyingpoint=i;
else if(stockvalues[i]>buy)
sell=stockvalues[i];
sellingpoint=i;
currentprofit=sell-buy;
if(profit<currentprofit&&sellingpoint>buyingpoint)
finalbuy=buy;
finalsell=sell;
profit=currentprofit;
if(profit>0)
System.out.println("Buy shares at "+finalbuy+" INR and Sell Shares "+finalsell+" INR and Profit of "+profit+" INR");
else
System.out.println("Don't do Share transacations today");
【讨论】：
【参考方案12】：确定最大利润的一种可能性可能是在数组中的每个索引处跟踪数组中的左侧最小和右侧最大元素。然后，当您遍历股票价格时，对于任何一天，您都会知道当天的最低价格，并且您还将知道当天之后（包括）当天的最高价格。
例如，让我们定义一个min_arr 和max_arr，给定数组是arr。 min_arr 中的索引 i 将是 arr 中所有索引 <= i （包括 i 的左侧）的最小元素。 max_arr 中的索引 i 将是 arr 中所有索引 >= i 中的最大元素（包括 i 的右侧）。然后，您可以找到max_arr 和 `min_arr' 中对应元素之间的最大差异：
def max_profit(arr)
min_arr = []
min_el = arr.first
arr.each do |el|
if el < min_el
min_el = el
min_arr << min_el
else
min_arr << min_el
end
end
max_arr = []
max_el = arr.last
arr.reverse.each do |el|
if el > max_el
max_el = el
max_arr.unshift(max_el)
else
max_arr.unshift(max_el)
end
end
max_difference = max_arr.first - min_arr.first
1.upto(arr.length-1) do |i|
max_difference = max_arr[i] - min_arr[i] if max_difference < max_arr[i] - min_arr[i]
end
return max_difference
end
这应该在 O(n) 时间内运行，但我相信它会占用大量空间。
【讨论】：
【参考方案13】：这是数组中两个元素之间的最大差异，这是我的解决方案：
O(N) 时间复杂度 O(1) 空间复杂度
int[] arr = 5, 4, 6 ,7 ,6 ,3 ,2, 5;
int start = 0;
int end = 0;
int max = 0;
for(int i=1; i<arr.length; i++)
int currMax = arr[i] - arr[i-1];
if(currMax>0)
if((arr[i] -arr[start])>=currMax && ((arr[i] -arr[start])>=(arr[end] -arr[start])))
end = i;
else if(currMax>(arr[i] -arr[start]) && currMax >(arr[end] - arr[start]))
start = i-1;
end = i;
max = arr[end] - arr[start];
System.out.println("max: "+max+" start: "+start+" end: "+end);
【讨论】：
【参考方案14】：在 FB 解决方案工程师职位的实时编码考试中失败后，我不得不在平静凉爽的气氛中解决它，所以这是我的 2 美分：
var max_profit = 0;
var stockPrices = [23,40,21,67,1,50,22,38,2,62];
var currentBestBuy = 0;
var currentBestSell = 0;
var min = 0;
for(var i = 0;i < (stockPrices.length - 1) ; i++)
if(( stockPrices[i + 1] - stockPrices[currentBestBuy] > max_profit) )
max_profit = stockPrices[i + 1] - stockPrices[currentBestBuy];
currentBestSell = i + 1;
if(stockPrices[i] < stockPrices[currentBestBuy])
min = i;
if( max_profit < stockPrices[i + 1] - stockPrices[min] )
max_profit = stockPrices[i + 1] - stockPrices[min];
currentBestSell = i + 1;
currentBestBuy = min;
console.log(currentBestBuy);
console.log(currentBestSell);
console.log(max_profit);
【讨论】：
不鼓励只回答代码。【参考方案15】：真正回答问题的唯一答案是@akash_magoon 之一（并且以如此简单的方式！），但它不会返回问题中指定的确切对象。我进行了一些重构，并让我在 php 中的答案返回了所要求的内容：
function maximizeProfit(array $dailyPrices)
$buyDay = $sellDay = $cheaperDay = $profit = 0;
for ($today = 0; $today < count($dailyPrices); $today++)
if ($dailyPrices[$today] < $dailyPrices[$cheaperDay])
$cheaperDay = $today;
elseif ($dailyPrices[$today] - $dailyPrices[$cheaperDay] > $profit)
$buyDay = $cheaperDay;
$sellDay = $today;
$profit = $dailyPrices[$today] - $dailyPrices[$cheaperDay];
return [$buyDay, $sellDay];
【讨论】：
【参考方案16】：一个巧妙的解决方案：
+ (int)maxProfit:(NSArray *)prices
int maxProfit = 0;
int bestBuy = 0;
int bestSell = 0;
int currentBestBuy = 0;
for (int i= 1; i < prices.count; i++)
int todayPrice = [prices[i] intValue];
int bestBuyPrice = [prices[currentBestBuy] intValue];
if (todayPrice < bestBuyPrice)
currentBestBuy = i;
bestBuyPrice = todayPrice;
if (maxProfit < (todayPrice - bestBuyPrice))
bestSell = i;
bestBuy = currentBestBuy;
maxProfit = (todayPrice - bestBuyPrice);
NSLog(@"Buy Day : %d", bestBuy);
NSLog(@"Sell Day : %d", bestSell);
return maxProfit;
【讨论】：
【参考方案17】：def get_max_profit(stock):
p=stock[0]
max_profit=0
maxp=p
minp=p
for i in range(1,len(stock)):
p=min(p,stock[i])
profit=stock[i]-p
if profit>max_profit:
maxp=stock[i]
minp=p
max_profit=profit
return minp,maxp,max_profit
stock_prices = [310,315,275,295,260,270,290,230,255,250]
print(get_max_profit(stock_prices))
python3中的这个程序可以返回最大化利润的买入价和卖出价，计算方法为O(n)的时间复杂度和O(1)的空间复杂度强>.
【讨论】：
【参考方案18】：这是我的解决方案
public static int maxProfit(List<Integer> in)
int min = in.get(0), max = 0;
for(int i=0; i<in.size()-1;i++)
min=Math.min(min, in.get(i));
max = Math.max(in.get(i) - min, max);
return max;
【讨论】：
【参考方案19】：对于多次买入/卖出， 使用下面的代码
时间复杂度 O(n)
n=int(input())
a = list(map(int,input().split()))
m=0
b=a[0]
s=0
for i in range(1,n):
if(a[i]<a[i-1]):
s=a[i-1]
m=m+s-b
b=a[i]
if(a[n-1]>b):
m=m+a[n-1]-b
print(m)
【讨论】：
【参考方案20】：对于跟踪最小和最大元素的所有答案，该解决方案实际上是 O(n^2) 解决方案。这是因为最后必须检查最大值是否出现在最小值之后。如果没有，则需要进一步迭代，直到满足该条件，这留下了 O(n^2) 的最坏情况。如果您想跳过额外的迭代，则需要更多空间。无论哪种方式，与动态规划解决方案相比都是禁忌
【讨论】：
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