Arrays 在股票价值数组中查找买入/卖出价格，以最大化正差异
在今天的一次采访中，我得到了这个问题，而它的优化解决方案让我觉得很冷（这很糟糕，因为我真的很想为这家公司工作……） 给定一个真实值数组，每个数组代表一家公司在任意时间段后的股票价值，找出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：Arrays 在股票价值数组中查找买入/卖出价格，以最大化正差异,arrays,optimization,complexity-theory,stocks,Arrays,Optimization,Complexity Theory,Stocks,在今天的一次采访中，我得到了这个问题，而它的优化解决方案让我觉得很冷（这很糟糕，因为我真的很想为这家公司工作……） 给定一个真实值数组，每个数组代表一家公司在任意时间段后的股票价值，找出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 需要注意的是，数组是临时“排序”的，即随着时间的推移，值会附加到数组的右端。因此，我们的买入价值将（必须）在卖出价值的左边 （在上述示
55.39 109.23 48.29 81.59 105.53 94.45 12.24
48.29105.53//返回一个2元素数组：第一个元素是参数数组中的索引
//第二个因素是最佳购买价格指数
//总的来说，使交易回报最大化的销售价格
//
//如果没有有利的交易（如价格单调下跌），则返回null
公共整数[]最大化回报（ArrayList价格）{
int[]retval=新的int[2]；
整数买入=0，卖出=1；
retval[BUY]=retval[SELL]=-1；//分别为买入和卖出价格指数
对于（int i=0；i0.0）{
如果（retval[BUY]<0 | | difference>prices.get（retval[SELL]）.doubleValue（）-
prices.get（retval[BUY]）.doubleValue（））{
归还[购买]=我；
retval[SELL]=j；
}
}
}
}
退货（退货[购买]>0？退货：空）；
}
#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;
}
#包括
int main（）
{
双倍真大号=1e99；
double-bottom=REALLY\u BIG\u NO；//任意大数
双货币百思买=0.0；
双层顶=0.0；
双倍现金最佳销售=0.0；
双倍利润=0.0；
//一系列价格
双倍价格[]={10.50,55.39,109.23,48.29,81.59,105.53,94.45,12.24,152.0,2170.0}；
int numPrices=10；//价格数量
对于（整数i=0；i顶部）
{
top=价格[i]；
//计算利润
双重潜在利润=（上下）；
如果（潜在利润>利润）&&
底部！=非常大（不）
{
利润=潜在利润；
currBestSell=顶部；
currBestBuy=底部；
}
}
}
std：：C#中的cout
static void Main（字符串[]args）
{
double[]值=新的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值[i-1]）
{
//向上趋势，附加到现有差异
diff+=值[i]-值[i-1]；
}
其他的
{
//向下，重置差异
差异=0；
}
如果（差异>最大差异）
{
maxDiff=diff；
最大值=数值[i]；
}
}
WriteLine（“在{0}买入；在{1}卖出”，max-maxDiff，max）；
}
编辑：基于@Joe失败的测试用例的新算法——顺便说一句，这和@Doug T现在的答案是一样的
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);
}
static void Main（字符串[]args）
{
double[]值=新的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；
双底=值[0]；
for（int i=1；i最大差异）
{
maxDiff=diff；
最大值=数值[i]；
}
if（值[i]<底部）
{
底部=数值[i]；
差异=0；
}
}
WriteLine（“在{0}买入；在{1}卖出”，max-maxDiff，max）；
}
我真的必须指出，作为一个面试问题，期望你解决它，因为O（n）是近乎荒谬的。面试问题是为了证明你能解决一个问题，而你能解决它。你在O（n^2）和O（n）中解决它的事实应该是无关紧要的。如果一家公司因为你没有在O（n）中解决这个问题而放弃聘用你这可能不是你想在的公司。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的：
（1） 对于每一天，如果我必须在那一天卖掉我的股票，我可以支付的最低购买金额是多少？必要的
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);
}
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);
}
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) + ")");
}
#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 );
}
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);
}
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]
}
//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);
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 );
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;
}
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!!);
}
}
def getBestProfit(prices):
lo = hi = profit = 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
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 );
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)
Best buy: 48.290000; Best sell: 105.530000
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
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;
}
}}
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;
}
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
# 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
min = 100000000
max = 0
for elem in inp:
if elem < min:
min = elem
tempMax = elem-min
if tempMax > max:
max = tempMax
print(max)
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));
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]));
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
}
}