這篇文章，會帶著大家複習以前學過的FSM+DP框架，
並且以有限狀態機 + DP狀態轉移的概念為核心，
貫穿一些相關聯最佳股票買賣系列的題目，
透過框架複現來幫助讀者理解這個實用的演算法框架。
基本的FSM + DP 框架，配合交易邏輯。
針對每一天，其實歸根究柢只有兩種狀態。
第一種是手上都持有現金。
第二種是手上持有股票。
每天進行的動作，也只有兩種可能情況:
假如今天結束的時候，是都持有現金。
代表昨天就已經持有現金今天繼續持有；或者，昨天持有股票，今天選擇賣出股票。
假如今天結束的時候，是持有股票。
代表昨天就已經持有股票今天繼續抱著；或者，昨天持有現金，今天選擇買入股票。
寫成虛擬碼或者演算法的話，可以這樣表達
DP[day_d, KeepCash]
= max( DP[day_d-1, KeepCash], DP[day_d-1, HoldStack] + stockPrice[day_d] )
DP[day_d, HoldStock]
= max( DP[day_d-1, HoldStock], DP[day_d-1, KeepCash] - stockPrice[day_d] )
接下來，我們會用這個上面這種結構，貫穿一些同類型，有關聯的題目
(請讀者、或觀眾留意上面的這種 FSM+ DP 動態規劃 框架，之後的解說會反覆出現)
FSM有限狀態機 + DP 動態規劃框架 示意圖
打鐵趁熱，現在來看最佳股票買賣系列題第一題。
Best time to buy and sell stock
這題的限制條件是只能做一組交易(也就是一次買入，一次賣出)。
只有一次買入，相當於會從某一天選擇買入，之後就只能選另一天賣出了。
某一天買入之前一定是沒賺沒賠，所以寫成式子就是 0 - stock_price
最後用FSM+DP框架的思考邏輯寫成程式碼，就是下方這個樣子
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# It is impossible to have stock to sell on first day, so -infinity is set as initial value
cur_hold, cur_cash = -float('inf'), 0
for stock_price in prices:
prev_hold, prev_cash = cur_hold, cur_cash
# either keep in hold, or just buy today with stock price
cur_hold = max(prev_hold, 0-stock_price)
# either keep in cash, or just sell today with stock price
cur_cash = max(prev_cash, prev_hold + stock_price)
# max profit must be in Cash state
return cur_cash
接著來看變化題，最佳股票買賣系列題第二題。
Best time to buy and sell stock II
這題比較寬鬆，不限制交易組數，也就是可以交易任意多組，
買入股票n次，賣出股票n次。
也就是說，某一天才買入的狀態會相依於前一天現金部位的狀態
所以寫成式子就是 dp[day-1, KEEP_CASH] - stock_price
最後用FSM+DP框架的思考邏輯寫成程式碼，就是下方這個樣子
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# Use constant literal to help reader understand source code below.
HOLD_STOCK, KEEP_CASH = 0, 1
## Dictionary (Hash table)
# key: (day, state) pair
# value: coreesponding maximal profit
dp = {}
# No free lunch, it is impoosible to have stock before first trading day
dp[-1, HOLD_STOCK] = -math.inf
# No gain, no loss before first trading day
dp[-1, KEEP_CASH] = 0
# For each day with corresponding stock price in stock market
for day, stock_price in enumerate(prices):
# If today we have stock, either we already had it yesterday, or we just buy stock and hold it today.
dp[day, HOLD_STOCK] = max( dp[day-1, HOLD_STOCK], dp[day-1, KEEP_CASH] - stock_price)
# If today we keep cash, either we already kept cash yesterday, or we just sell out stock today
dp[day, KEEP_CASH] = max( dp[day-1, KEEP_CASH], dp[day-1, HOLD_STOCK] + stock_price)
#---------------------------------------------------------------
# To get maximal realized profit, final state must be KEEP_CASH.
last_day = len(prices)-1
return dp[last_day, KEEP_CASH]
Best time to buy and sell stock II 空間優化的程式碼實作
仔細觀察，可以發現每天的狀態只依賴前一天的狀態去更新。
因此，可以用之前學過的狀態壓縮的技巧，把表格拿掉，
改成只用變數儲存狀態，節省空間，從O(n)空間降到O(1)常數級別。
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# It is impossible to sell stock on first day, set -infinity as initial value for cur_hold
cur_hold, cur_cash = -float('inf'), 0
for stock_price in prices:
prev_hold, prev_cash = cur_hold, cur_cash
# either keep hold, or buy in stock today at stock price
cur_hold = max( prev_hold, prev_cash - stock_price )
# either keep cash, or sell out stock today at stock price
cur_cash = max( prev_cash, prev_hold + stock_price )
# maximum profit must be in Cash state
return cur_cash
有了狀態圖以後，再回來看程式碼清晰許多了，對吧!
接下來，為了再更逼真一點，延伸題就是把交易的額外成本fee(稅, 手續費)也加入考量。
延伸題就是
Best time to buy and sell stock with transaction fee
那也很容易，我們就在每組交易進行的時候，額外多扣一個費用fee即可。
註: 不失一般性，我們選擇在買入股票時付出額外的交易成本費用fee。
讀者想要在賣出時扣掉也可以，可以試著改改看，也能通過測試！
Best time to buy and sell stock with transaction fee 程式碼實作
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
cur_hold, cur_cash = -math.inf, 0
for stock_price in prices:
prev_hold, prev_cash = cur_hold, cur_cash
# either keep cash, or sell out today at stock price
cur_cash = max(prev_cash, prev_hold + stock_price )
# either keep hold, or buy in today at stock price and pay transaction fee for this trade
cur_hold = max(prev_hold, prev_cash - stock_price - fee)
# maximum profit must be in not-hold state
return cur_cash
好，接著來看進階題，最佳股票買賣系列題第三題。
Best time to buy and sell stock III
這題的限制條件是最多只能兩組交易(也就是各兩次買入，各兩次賣出)。
也就是說，第i次的買入並且持有股票的狀態，
前面會相依於第(i-1)次賣出並且持有現金的狀態。
所以寫成式子就是 dp[KEEP_CASH, i-1] - stock_price
最後用FSM+DP框架的思考邏輯寫成程式碼，就是下方這個樣子
class Solution:
def maxProfit(self, prices: List[int]) -> int:
HOLD_STOCK, KEEP_CASH = 0, 1
# dictionary
# key: state, kth-transaction
# value: corresponding profit
dp = defaultdict(int)
# No free lunch, impoosible to have stock before first trading day
dp[(HOLD_STOCK, 0)] = -math.inf
dp[(HOLD_STOCK, 1)] = -math.inf
dp[(HOLD_STOCK, 2)] = -math.inf
for stock_price in prices:
## For 1st transaction:
# Either we kept cash already, or we just sell out stock today
dp[KEEP_CASH, 1] = max(dp[KEEP_CASH, 1], dp[HOLD_STOCK, 1] + stock_price )
# Either we had stock already, or we just buy in stock today ( add one more transaction)
dp[HOLD_STOCK, 1] = max(dp[HOLD_STOCK, 1], dp[KEEP_CASH, 0] - stock_price)
## For 2nd transaction:
# Either we kept cash already, or we just sell out stock today
dp[KEEP_CASH, 2] = max(dp[KEEP_CASH, 2], dp[ HOLD_STOCK, 2] + stock_price)
# Either we had stock already, or we just buy in stock today ( add one more transaction)
dp[HOLD_STOCK, 2] = max(dp[HOLD_STOCK, 2], dp[KEEP_CASH, 1] - stock_price)
# Maximal profit must be KEEP_CASH on last day
# (This means we cash out and sell stocks finally)
return dp[KEEP_CASH, 2]
仔細觀察，其實for loop裡面包含著一個階梯迭代更新的關係。
第i組的交易狀態更新，只依賴第i組和第i-1組的狀態。
抽象化+結構化，可以寫成這個形式
for trans in range(1, k+1):
dp[HOLD_STOCK, trans] = max( dp[HOLD_STOCK, trans], dp[KEEP_CASH, trans-1 ] - stock_price )
dp[KEEP_CASH, trans] = max( dp[KEEP_CASH, trans], dp[HOLD_STOCK, trans] + stock_price )
寫成這個形式，也相當於推廣成
最多只能k組交易下的最佳股票買賣模式
當我們推導出這個通則的時候，其實無形中已經把下一題也解出來了！ XD
下一題就是
Best time to buy and sell stock IV
這題的限制條件是最多只能k組交易(也就是各k次買入，各k次賣出)。
這邊承接第剛剛上面推導出來的通則，順著寫下來即可。
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
n = len(prices)
HOLD_STOCK, KEEP_CASH = 0, 1
dp = defaultdict(int)
for k in range (0, k+1):
dp[HOLD_STOCK, k] = -math.inf
for stock_price in prices:
for trans in range(1, k+1):
dp[HOLD_STOCK, trans] = max( dp[HOLD_STOCK, trans], dp[KEEP_CASH, trans-1 ] - stock_price )
dp[KEEP_CASH, trans] = max( dp[KEEP_CASH, trans], dp[HOLD_STOCK, trans] + stock_price )
return dp[KEEP_CASH, k]
萬變不離其宗，其實DP狀態轉移的觀念其實都是相通的。
最後一題是一個延伸變化題
Best Time to Buy and Sell Stock with Cooldown
這題的也是不限制交易組數，也就是可以交易任意多組，
買入股票n次，賣出股票n次。
也就是說，某一天才買入的狀態會相依於前一天現金部位的狀態
但是，多了一個冷卻條件，
股票賣出之後，會強迫進入冷卻日一天，冷卻日的次日才可以再買入股票。
畫成FSM有限狀態機的狀態圖就是
順著狀態圖的更新邏輯寫下迭代耕新的程式碼即可
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# initialization
cool_down, sell, hold = 0, 0, -float('inf')
for stock_price_of_Day_i in prices:
prev_cool_down, prev_sell, prev_hold = cool_down, sell, hold
# Max profit of cooldown on Day i comes from either cool down of Day_i-1, or sell out of Day_i-1 and today Day_i is cooling day
cool_down = max(prev_cool_down, prev_sell)
# Max profit of sell on Day_i comes from hold of Day_i-1 and sell on Day_i
sell = prev_hold + stock_price_of_Day_i
# Max profit of hold on Day_i comes from either hold of Day_i-1, or cool down on Day_i-1 and buy on Day_i
hold = max(prev_hold, prev_cool_down - stock_price_of_Day_i)
# The action of final trading day must be either sell or cool down
return max(sell, cool_down)
可以視為Best time to buy and sell stock II的延伸題(多了額外的冷卻條件)
結語
好，今天一口氣介紹了最精華的部分，
通用的FSM有限狀態機 + DP狀態轉移框架，還有相關的衍伸變化題與演算法建造流程。
希望能幫助讀者、觀眾徹底理解它的原理，
並且能夠舉一反三，洞察其背後的本質和了解相關的應用領域！
感謝收看囉! 我們下篇文章再見~
