股市的交易日(動態規劃演算法)
阿新 • • 發佈:2018-12-07
股市的交易日
///
/// 在股市的交易日中,假設最多可進行兩次買賣(即買和賣的次數均小於等於2),規則是必須一筆成交後進行另一筆(即買-賣-買-賣的順序進行)。給出一天中的股票變化序列,請寫一個程式計算一天可以獲得的最大收益。
///給定價格序列prices及它的長度n,請返回最大收益。保證長度小於等於500。
///測試樣例:
///[10,22,5,75,65,80],6
///返回:
///87
///
想法:
最多兩次交易,前一次交易與後一次交易的結果想加,有關聯,所以選用 動態規劃演算法
把序列分為兩段
…
10 ||22 5 75 65 80
10 22||5 75 65 80
10 22 5||75 65 80
…
計算前一段跟後一段的最優解相加最優
c# 程式碼1
///使用:
/// int[] prices = { 10,2,5,75,65,80,5,6 };
/// int max = maxProfit2(prices, prices.Length);
///方法:
static int maxProfit2(int[] prices, int n)
{
Console.WriteLine("maxProfit2");
int times = 0;
int max = 0;
int onemax = 0, twomax = 0;
for (int i = 0; i < n; i++)
{
onemax = 0; twomax = 0;
for (int j2 = 0; j2 < i; j2++)
{
for (int j3 = j2; j3 < i; j3++)
{
times++;
int vTmp = prices[j3] - prices[j2];
if (vTmp > onemax)
{
onemax = vTmp;
}
}
}
for (int k2 = i; k2 < n; k2++)
{
for (int k3 = k2; k3 < n; k3++)
{
times++;
int vTmp = prices[k3] - prices[k2];
if (vTmp > twomax)
{
twomax = vTmp;
}
}
}
Console.WriteLine(onemax + "--"+ twomax);
int tmpMax = onemax + twomax;
if (max < tmpMax)
{
max = tmpMax;
}
}
Console.WriteLine("maxProfit2-end:" + times);
return max;
}
c# 程式碼2
///使用:
/// int[] prices = { 10,2,5,75,65,80,5,6 };
/// int max = maxProfit3(prices, prices.Length);
///方法:
static int maxProfit3(int[] prices, int n)
{
Console.WriteLine("maxProfit3");
int max = 0;
int times = 0;
int onemax, twomax;
for (int i = 0; i < n; i++)
{
onemax = 0; twomax = 0;
for (int j = 0; j < i; j++)
{
for (int j2 = 0; j2 < j; j2++)
{
times++;
int tmpPri = prices[j] - prices[j2];
if (tmpPri > onemax)
{
onemax = tmpPri;
}
}
}
for (int k = i + 1; k < n; k++)
{
times++;
if ((prices[k] - prices[i]) > twomax)
{
twomax = prices[k] - prices[i];
}
}
Console.WriteLine(onemax + "--" + twomax);
int tmpMax = onemax + twomax;
if (tmpMax > max)
{
max = tmpMax;
}
}
Console.WriteLine("maxProfit3-end:" + times);
return max;
}