20

今天面试被问到这个问题，它的优化方案让我不寒而栗（吹爆了，因为我真的很想去这家公司工作……）

给定一组真实值，每个值代表任意时间段后公司的股票价值，找到最佳买入价及其对应的最佳卖出价（低买高卖）。

为了举例说明，让我们以 Z 公司的股票代码为例：

55.39 109.23 48.29 81.59 105.53 94.45 12.24

需要注意的重要一点是数组在时间上是“排序的”——即随着时间的推移，值被附加到数组的右端。因此，我们的买入价值将（必须）在我们的卖出价值的左侧。

（在上面的例子中，理想的解决方案是在 买入48.29和卖出105.53）

我用 O(n 2 ) 的复杂度（在 java 中实现）很容易地想出了简单的解决方案：

// returns a 2-element array: first element is the index in the argument array
// of the best buying price, and the second element is the index of the best
// selling price which, collectively, maximize the trading return
//
// if there is no favorable trading (e.g. prices monotonically fall), null is returned
public int[] maximizeReturn(ArrayList<Double> prices) {
  int [] retval = new int[2];
  int BUY = 0, SELL = 1;
  retval[BUY] = retval[SELL] = -1; // indices of buy and sell prices, respectively

  for (int i = 0; i < prices.size(); i++) {
    for (int j = i + 1; j < prices.size(); j++) {
      double difference = prices.get(j).doubleValue() - 
                          prices.get(i).doubleValue();

      if (difference > 0.0) {
        if (retval[BUY] < 0 || difference > prices.get(retval[SELL]).doubleValue() - 
                                            prices.get(retval[BUY]).doubleValue()) {
          retval[BUY] = i;
          retval[SELL] = j;
        }
      }
    }
  }
  return (retval[BUY] > 0 ? retval : null);
}

这就是我搞砸的地方：有一个线性时间 O(n) 解决方案，我完全试图弄清楚它（是的，我知道，失败）。有谁知道如何实现线性时间解决方案？（任何你喜欢的语言）谢谢！

编辑

我想，对于任何有兴趣的人，我今天刚收到消息说我没有得到我面试的工作，他们问我这个问题。:(

4

24 回答 24

14

在 C# 中：

static void Main(string[] args)
{
    double[] values = new double[7]{55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24};

    double max = double.MinValue, maxDiff = double.MinValue, diff = 0;

    for (int i = 1; i < values.Length; i++)
    {
        if (values[i] > values[i - 1])
        {
            //trending upward, append to existing differential
            diff += values[i] - values[i - 1];
        }
        else
        {
            //trending downward, reset the diff
            diff = 0;
        }

        if (diff > maxDiff)
        {
            maxDiff = diff;
            max = values[i];
        }
    }

    Console.WriteLine("Buy at {0}; Sell at {1}", max - maxDiff, max);
}

编辑：基于@Joe 失败的测试用例的新算法 - Nice Catch BTW！这也与@Doug T 现在的答案相同......

static void Main(string[] args)
{
    double[] values = new double[8] { 55.39, 109.23, 48.29, 81.59, 81.58, 105.53, 94.45, 12.24 };

    double max = double.MinValue, maxDiff = double.MinValue, diff = 0;
    double bottom = values[0];

    for (int i = 1; i < values.Length; i++)
    {
        diff += values[i] - values[i - 1];

        if (diff > maxDiff)
        {
            maxDiff = diff;
            max = values[i];
        }

        if (values[i] < bottom)
        {
            bottom = values[i];
            diff = 0;
        }
    }

    Console.WriteLine("Buy at {0}; Sell at {1}", max - maxDiff, max);
}
于 2009-11-02T20:52:50.247 回答
10

这是一个尝试（C++）。基本上每次我追踪一个新的顶部时，我都会尝试看看这是否是迄今为止最好的利润。我知道“底部”一定是更早发现的。那时我记得顶部、底部和当前的最大利润。如果稍后发现新的底部，它在当前顶部之后，所以我们必须重新设置顶部，看看稍微低一点的“顶部”是否可以产生更好的利润。

#include <iostream>

int main()
{

    double REALLY_BIG_NO = 1e99;
    double bottom = REALLY_BIG_NO; // arbirtrary large number
    double currBestBuy = 0.0;
    double top = 0.0;
    double currBestSell = 0.0;
    double profit = 0.0;

    // array of prices
    double prices[] = {10.50, 55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24, 152.0, 2, 170.0};
    int numPrices = 10;// number of prices

    for (int i = 0; i < numPrices; ++i)
    {
         if (prices[i] < bottom)
         {
            bottom = prices[i];
            // reset the search on a new bottom
            top = 0.0;
         }
         else if (prices[i] > top)
         {
            top = prices[i];
           // calculate profit
            double potentialProfit = (top - bottom);
            if (potentialProfit > profit &&
                bottom != REALLY_BIG_NO)
            {
                profit = potentialProfit;
                currBestSell = top;
                currBestBuy = bottom;
            }
         }
    }

    std::cout << "Best Buy: " << currBestBuy << "Best Sell: " << currBestSell << std::endl;
}

到目前为止，我已经玩过一堆不同的输入集，到目前为止我还没有遇到任何问题......（如果你测试这个并发现任何问题，请告诉我）

我强烈建议使用Austin Salonen对此问题的更新答案并将其调整为您的语言。

于 2009-11-02T20:48:50.090 回答
3

这个想法很简单。保留两个指针，lo 和 hi。
做一个Foo循环

  1. 如果价格高于 hi，更新 hi = price，继续
  2. 如果价格低于喜。那么 lo 和 hi 是可能的候选人之一。计算利润，如果它比以前的利润大，则存储它，然后将 lo, hi 重置为价格

def getBestProfit（价格）：
    lo = hi = 利润 = 0


    for price in prices:
        if lo == 0 and hi == 0:
            lo = hi = price

        if price > hi:
            hi = price

        if price < low:
            tmpProfit = hi - lo
            if tmpProfit > profit:
                profit = tmpProfit

            lo = hi = price
    return profit

就是这样

于 2013-09-12T01:41:53.557 回答
2 
void getBestTime (int stocks[], int sz, int &buy, int &sell){
int min = 0;
int maxDiff = 0;
buy = sell = 0;
for (int i = 0; i < sz; i++) 
{
    if (stocks[i] < stocks[min])
    {
        min = i;
    }
    int diff = stocks[i] - stocks[min];
    if (diff > maxDiff) 
    {
        buy = min;
        sell = i;
        maxDiff = diff;
    }
}}

以防万一您更喜欢这个答案。我在另一个网站上找到了它，但仍然如此。来源： http: //leetcode.com/2010/11/best-time-to-buy-and-sell-stock.html

于 2014-06-23T23:33:05.340 回答
1 
      public void profit(float stock[], int arlen ){
            float buy = stock[0];
            float sell = stock[arlen-1];
            int bi = 0;
            int si = arlen - 1;

            for( int i = 0; i < arlen && bi < si ; i++){

                    if( stock[i] <  buy && i < si){
                            buy = stock[i];
                            bi = i;
                    }
                    if(stock[arlen - i - 1] > sell &&  (arlen - i -1)  > bi){
                            sell = stock[arlen - i - 1];
                            si = arlen - i - 1;
                    }
            }
            System.out.println(buy+" "+sell);
    }
于 2011-07-06T09:08:01.627 回答
0

我真的必须指出，作为一个面试问题，希望你能解决它，因为 O(n) 是荒谬的。面试问题旨在证明您可以解决问题，并且您能够解决它。您在 O(N^2) 与 O(N) 中解决它的事实应该是无关紧要的。如果一家公司因为没有在 O(N) 中解决这个问题而放弃雇用你，那可能不是你无论如何都想工作的公司。

于 2009-11-02T21:59:53.223 回答
0

我想描述一下我是如何解决这个问题的，以便更容易理解我的代码：

(1) 对于每一天，如果我必须在那天卖出我的股票，我可以支付的最低购买金额是多少？本质上，我会在那天之前跟踪最低价格

(2) 对于每一天，如果我在那天卖出，我能赚多少钱？(当日股价-最低价)

这表明我必须跟踪两件事：（1）迄今为止的最低股价（2）迄今为止的最佳收益。

问题变成了选择哪一天出售。我会在能给我带来最好收入的那一天卖出。这是我的Java代码：

    public static void findBestDeal(double [] stocks) {
    double minsofar = stocks[0];
    double bestsofar = 0.0;

    for(int i=1; i< stocks.length; i++) {

        // What is the cheapest price to buy it if I'm going to sell it today
        if(stocks[i-1] < minsofar) {
            minsofar = stocks[i-1];
        }

        // How much do I earn if I sell it on ith day?
        double current_deal = stocks[i] - minsofar;

        // Is selling today better?
        if(current_deal > bestsofar) {
            bestsofar = current_deal;
        }
    }

    System.out.println("Found the best deal: " + bestsofar + " (Bought at " + minsofar + " and sold at " + (minsofar+bestsofar) + ")");

}
于 2011-02-18T10:43:12.547 回答
0

这是我的 O(n) 实现。我正在使用更改数组来计算最大利润和买卖日期。欢迎您对此发表评论。

#include<stdafx.h>
#include<stdio.h>

int main()
{
    //int arr[10] = {15, 3, 5,9,10,1,6,4,7,2};
    int arr[7] = {55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24};
    int change[7];
    int n=7;
    for(int i=1;i<=n;i++)
    {
    change[i] = arr[i]- arr[i-1];
    }
    int i=0,index = 0;
    int sum = 0;
    int maxsum = 0;
    int startpos = 0;
    int endpos = 0;
    while(index < n)
    {
        sum = sum + change[index];
        if(maxsum < sum)
        {
        maxsum = sum; 
        startpos = i;
        endpos = index;

        }
        else if (sum<0) // negative number ,set sum to zero
        {
        sum = 0;
        i=index+1;
        }
        index++;
    }

    printf("max profit is%d %d %d", maxsum , startpos, endpos+1 );
}
于 2011-03-24T00:03:27.017 回答
0

在我努力学习围棋的过程中，也为了在这个问题上绞尽脑汁，这是我的尝试。

func GetMaxProfit2(prices []float64) (float64, float64) {
    var min, max, pmin, pmax int

    for i, v := range prices {
        if v - prices[min] > prices[max] - prices[min] {
            pmax = max
            max = i
        }
        // Reset the max when min is updated.
        if v < prices[min] {
            pmin = min
            min = i
            pmax = max
            max = i
        }
    }

    // If min is ahead of max, reset the values back    
    if min >= max {
        min = pmin
        max = pmax
    }

    return prices[min], prices[max]
}
于 2011-12-23T06:43:53.163 回答
0

这是我使用 Javascript 的尝试。该脚本以 O(N) 计算答案：

//Main Stock Array
var stock = [15, 20, 0, 3, 30, 45, 67, 92, 1, 4, 99];


//Setup initial variable state
var ans = {}, tmp = {}; //These are just for namespacing / syntatic sugar
ans.minVal = stock[0];
ans.minInd = 0;
ans.maxDiff = stock[1] - stock[0];
ans.maxInd = 1;
tmp.minInd = ans.minInd;
tmp.minVal = ans.minVal;

//Basically we iterate throught the array. If we find a new low, we start tracking it. Otherwise we compare the current index against the previously found low
for(i = 1; i <= stock.length-1; i++) {
    if(tmp.minVal > stock[i]) {
        tmp.minVal = stock[i];
        tmp.minInd = i;
    } else {
        ans.diff = stock[i] - stock[tmp.minInd];
        if(ans.diff > ans.maxDiff) { //Looks like we found a new maxDifference. Lets log the indexes
            ans.maxDiff = ans.diff;
            ans.maxInd = i;
            ans.minInd = tmp.minInd;
            ans.minVal = tmp.minVal;
        }
    }
}

document.write('You should buy your stocks on day ' + ans.minInd + ' and sell on day ' + ans.maxInd);
于 2012-04-01T09:59:37.417 回答
0

这是一个实际有效的 C 解决方案：

无效 bestBuySell() { 双倍价格 [] = {10.50, 10.0, 3.0, 194.0, 55.39, 2.0, 109.23, 48.29, 81.59, 105.53, 94.45, 191.0, 200.0, 12.24}; int arrSize = 14; 双倍 bestBuy = 价格 [0]，bestSell = 价格 [1]，bestPotentialBuy = 价格 [0]；双潜在利润 = 价格[1] - 价格[0]；

for(int i = 1; i < (arrSize-1); i++)
{
    if(prices[i] < bestBuy)
        bestPotentialBuy = prices[i];            

    if((prices[i+1] - bestPotentialBuy) > potentialProfit)
    {
        bestBuy = bestPotentialBuy;
        bestSell = prices[i+1];
        potentialProfit = prices[i+1] - bestPotentialBuy;
    }
}

printf( "bestBuy %f bestSell %f\n", bestBuy, bestSell );

}

于 2013-02-20T22:42:23.120 回答
0

1.我们不能简单地将值中的最小金额作为“Best Buy”，将最大金额作为“Best Sell”，因为“Sell”必须在“Buy”之后发生。

2.我们不能将记录的最低值视为“Best Buy”，因为随后几天的股票价值可能与记录的最低值的差异可能产生的利润可能低于“记录的利润”。

3.Best Buy 和 Best Sell 被视为单一变体，因为正是这些值之间的正差使利润最大化。

4.由于过去任何记录的最低值都是潜在的买入候选者，因此必须始终对照记录的最低值和当天的股价检查最大利润条件。因此，我们始终必须跟踪记录的最低值，但只是存在由于原因 2，记录的最小值不构成“百思买”。

现在让下面的代码在 O(n) 次中执行将是有意义的。

public class BestStockBuyAndSell {

public static void main(String[] args) {

    double[] stockPrices = {55.39,109.23,48.29,81.59,105.53,94.45,12.24};
    int [] bestBuySellIndex = maxProfit(stockPrices);

    System.out.println("Best Buy At "+stockPrices[bestBuySellIndex[0]]);
    System.out.println("Best Sell At "+stockPrices[bestBuySellIndex[1]]);

    System.out.println("Max Profit = "+(stockPrices[bestBuySellIndex[1]]-stockPrices[bestBuySellIndex[0]]));

}

public static int[] maxProfit(double[] stockPrices)
{
    int bestBuy=0;
    int bestSell=0;

    int[] bestCombination ={bestBuy,bestSell};
    double recordedMinimum = stockPrices[bestBuy];
    int recordedMinimuIndex = bestBuy;
    double bestProfitSofar = stockPrices[bestSell] - stockPrices[bestBuy];

    for(int i=1;i<stockPrices.length;i++)
    {
        if(stockPrices[i] - recordedMinimum > bestProfitSofar)
        {

            bestProfitSofar = stockPrices[i] - recordedMinimum;
            bestSell = i;
            bestBuy = recordedMinimuIndex;
        }

        if(stockPrices[i] < recordedMinimum)
        {
            recordedMinimuIndex = i;
            recordedMinimum = stockPrices[i];
        }

    }

    bestCombination[0] = bestBuy;
    bestCombination[1] = bestSell;


    return bestCombination;

}

}

于 2013-04-01T13:37:49.897 回答
0

我为这个问题想出了以下算法，似乎适用于所有输入。此外，如果 Stock 值持续下降，程序将输出不购买该股票：

  public class GetMaxProfit 
  { 

  double minValue = -1, maxValue = -1;
  double maxDiff = 0;

  public void getProfit(double [] inputArray){
    int i=0, j=1;
    double newDiff = 0;
    while(j<inputArray.length){
         newDiff = inputArray[j]-inputArray[i];
         if(newDiff > 0){
             if(newDiff > this.maxDiff){
               this.minValue = inputArray[i];
               this.maxValue = inputArray[j];
               this.maxDiff = newDiff;
             }
        }
        else{
            i = j;
        }
        j++;
    }
 }

 public static void main(String[] args) {
    // TODO Auto-generated method stub
    GetMaxProfit obj = new GetMaxProfit();

    obj.getProfit(new double[]{55.39, 19.23, 14.29, 11.59, 10.53, 9.45, 1.24});
    if(obj.minValue != -1 && obj.maxValue != -1){
      System.out.println("Buy Value for the input: "+obj.minValue);
      System.out.println("Sell Value for the input: "+obj.maxValue);
      System.out.println("Best profit for the input: "+obj.maxDiff);
            }
            else
               System.out.println("Do Not Buy This STOCK!!);

 }

}

你有什么可以找到的吗？它的时间复杂度是 O(N)

于 2013-07-19T18:43:07.033 回答
0

这是我的解决方案，与@Doug T 相同。除了我还在索引中跟踪当天。请提供反馈。

 int prices[] = {4,4,5,6,2,5,1,1};
 //int prices[] = {100, 180, 260, 310, 40, 535, 695};

 int currentBestSellPrice=0;
 int currentBestBuyPrice=0;
 int lowindex=0;
 int highindex=0;
 int low=prices[0];
 int high=prices[0];
 int profit=0;
 int templowindex=0;
 for(int i=0; i< prices.length;i++)
 {
     // buy low
     if(prices[i] < low && i+1 < prices.length)
     {
         low = prices[i];  
         templowindex=i;
         high=0;
     }
     // sell high
     else if(prices[i] > high)
     {
         high = prices[i];
         int potentialprofit = (high-low);
         if(potentialprofit > profit)
         {
             profit = potentialprofit;
             currentBestSellPrice = high;
             currentBestBuyPrice = low;
             highindex=i;
             lowindex=templowindex;
         }
     }
 }


 System.out.println("Best Buy Price : "+ currentBestBuyPrice + " on day "+ lowindex);
 System.out.println("Best Sell Price : "+ currentBestSellPrice+ " on day "+ highindex );
于 2014-01-27T20:18:02.667 回答
0

F# 解决方案适用于那些对功能感兴趣的人。我不会说它有那么大的不同。

let start, _, profit = 
    [55.39; 109.23; 48.29; 81.59; 81.58; 105.53; 94.45; 12.24 ]
    |> Seq.fold (fun (start,newStart,profit) i -> 
                    let start = defaultArg start i
                    let newStart = defaultArg newStart i
                    let newProfit = i - newStart
                    if profit < newProfit 
                    then  Some newStart, Some newStart,newProfit
                    else if start > i 
                    then Some start, Some i, profit 
                    else Some start,Some newStart,profit) (None,None, 0.0)
printf "Best buy: %f; Best sell: %f" start.Value (start.Value + profit)

输出：

Best buy: 48.290000; Best sell: 105.530000
于 2014-02-10T19:45:42.233 回答
0

这是我在 Ruby 中的解决方案：

values = [55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24]

max_diff = 0
diff = 0
min = values[0]
max = 0

values.each_with_index do |value, index = 1|
  # get an array of the previous values before the current one
  lag_values = values[0..index]

  # get the minimum of those previous values
  min_lag_value = lag_values.min

  # difference between current value and minimum of previous ones
  diff = values[index].to_i - min_lag_value.to_i

  # if current difference is > previous max difference, then set new values for min, max_diff, and max
  if diff > max_diff
    max_diff = diff
    min = min_lag_value
    max = values[index]
  end
end

min # => 48.29
max # => 105.3
max_diff # => 57

干杯

于 2014-05-14T18:20:57.090 回答
0

我得到了相同的100％，给你。

public int solution(int[] A) {
      if (A == null || A.length<=1){
            return 0;
        }
        int minValue = Math.min(A[0], A[1]);
        int profit = A[1] - A[0];
        for (int i = 2; i < A.length; i++) {
          minValue = Math.min(minValue, A[i]);
          profit = Math.max(A[i] - minValue,profit);
        }

        return profit > 0 ? profit : 0;
}
于 2014-11-11T11:03:34.430 回答
0

我的想法是，对于每个指数i，出售这只股票的理想指数是什么？这当然是 之后的最大值的索引i。i这将我们的问题减少到在每个iin 的索引之后找到最大元素[1 ... n]如果我们能O(n)及时做到这一点，那么我们可以在其中找到最佳选择并报告它。

一种方法是从数组的末尾开始遍历，维护两个变量，一个保存我们迄今为止遇到的最大元素，一个保存我们到目前为止max_till_now可以获得的最大利润，profit. 这只是为了让我们可以一次性完成。我们还可以使用额外的空间，并为每个元素i存储范围内最大元素的索引，[i + 1 ... n]然后找到最大利润。

这是我的python代码：

def buyLowSellHigh(L):
    length = len(L)
    profit = 0
    max_till_now = L[length - 1]
    for i in xrange(length - 2, -1, -1):
        if L[i] > max_till_now: max_till_now = L[i]
        else:
            if max_till_now - L[i] > profit: profit = max_till_now - L[i]
    return profit
于 2014-11-13T06:55:05.147 回答
0

另一个 Ruby 解决方案：

# Here's some examples. Please feel free to give your new test.
values = [55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24]
# values = [5, 6, 4, 7, 9, 8, 8]
# values = [5, 10, 4, 6, 7]
# values = [5, 10, 4, 6, 12]
# values = [1, 2, 3, 4, 5]



# Initialize parameters.
min = values[0]
best_buy_time = values[0]
best_sell_time = values[0]
max_profit = 0



# This solution is based on comparing previous k elements and k+1 one.
# The runtime is O(n) and it only use O(1) auxiliary storage.
values.each_with_index do |value, index = 1|

  # Check value in this turn.
  puts value

  # Check current value is bigger than min or not.
  # If not, we find the new min.
  if value <= min
    min = value

  # If current value is bigger than min and
  # (value - min) is bigger than previous max_profit,
  # set new best_buy_time, best_sell_time & max_profit.
  else
    if value - min >= max_profit
      best_buy_time = min
      best_sell_time = value
      max_profit = value - min
    end

  end

end



# Let's see about the result.
puts "\nbest_buy_time: ", best_buy_time, "\nbest_sell_time: ", best_sell_time, "\nmax_profit: ", max_profit
于 2015-09-16T01:32:27.120 回答
0

那这个呢？

min = 100000000
max = 0

for elem in inp:
    if elem < min:
       min = elem
    tempMax = elem-min
    if tempMax > max:
        max = tempMax

print(max)
于 2015-12-29T16:40:08.300 回答
0

javascript中的解决方案

var stockArr = [13931, 9889, 987, 4, 89, 100];

function getBestTime(sortedArr) {
  var min = 0;
  var buyIndx = 0;
  var saleIndx = 0;
  var maxDiff = 0;
  for (var i = 0; i < stockArr.length; i++) {
    if (stockArr[i] < stockArr[min]) {
      min = i;
    }
    var diff = stockArr[i] - stockArr[min];
    if (diff > maxDiff) {
      buy = min;
      sale = i;
      maxDiff = diff;
    }
  }
  return {
    buy:buy+1,
    sale:sale+1,
    diff:maxDiff
  }
}

console.log(getBestTime(stockArr));
于 2016-05-26T17:21:52.587 回答
0

这是一个javascript解决方案..

function getMax(arr){
        //we need more than at least 3 ints to be able to do this
        if(arr.length <= 1) return arr;
        // get the minimum price we can sell at to make a profit
        var min = arr[0];
        //get the first potential maximum profit
        var max = arr[1] - arr[0];

        //while looping through we must get a potential value, 
       //we can then compare that using the math.max using the maximum
      //and the potential prices that we have seen. Once the loop runs the ouput here should be 6!
        for(var i = 1; i < arr.length; ++i){
            var current = arr[i];
            var potential = current - min;

            max = Math.max(max, potential);
            min = Math.min(min, current);
        }

        return max;
    }

    console.log(getMax([10, 7, 5, 8, 11, 9]));

运行时间为 O(n)

于 2016-07-06T09:24:29.490 回答
0

scala 中的解决方案：

示例：[ 7、2、5、6、1、3、6、4]

保留一个指向最后最低股价（lastStockPrice）的指针，并将其与当前股价进行比较。当您达到当前股票价格 < 最后最低股票价格的点时，您更新 lastStockPrice。

在遍历数组时，请跟踪 currentPrice 和 lastStockPrice 之间的最大差异（利润），因为更新 lastStockPrice 时利润可能会发生变化。

下面的 scala 代码在 O(n) 时间内工作并占用恒定的空间。

object Solution {
    def maxProfit(prices: Array[Int]): Int = {
        var lastStockPrice = Int.MaxValue
        var maxProfit = 0
        for(currentPrice <- prices){
            if(currentPrice < lastStockPrice){
                lastStockPrice = currentPrice;
            }else if(currentPrice - lastStockPrice > maxProfit){
                maxProfit = currentPrice - lastStockPrice;
            }
        }
        maxProfit
    }
}
于 2018-04-28T03:00:02.033 回答
0

解决这个问题的逻辑与使用 Kadane 算法的“最大子数组问题”相同。由于到目前为止还没有任何机构提到这一点，我认为每个人都知道这是一件好事。

所有直截了当的解决方案都应该有效，但是如果面试官通过给出不同的价格数组来稍微扭曲问题，例如：对于 {1, 7, 4, 11}，如果他给出 {0, 6, -3, 7} ，你最终可能会感到困惑。

这里的逻辑是计算原始数组的差值（maxCur += prices[i] - prices[i-1]），并找到一个连续的子数组给出最大的利润。如果差值低于 0，则将其重置为零。

class Solution:
def maxProfit(self, prices: List[int]) -> int:
    
    _currmax = 0
    _globalMax = 0
    
    for i in range(1,len(prices)):
        
        _currmax = max(_currmax+(prices[i]-prices[i-1]),0)
        _globalMax = max(_globalMax,_currmax)
        
    return _globalMax
于 2021-07-06T17:24:18.383 回答