- 121. Best Time to Buy and Sell Stock：股票一次買入賣出，求最大利潤
class Solution(object):
def maxProfit(self, prices):
if prices == []: return 0
minNum,ret = prices[0],0
for p in prices:
minNum = min(minNum,p)
ret = max(ret,p-minNum)
return ret- 122. Best Time to Buy and Sell Stock II：股票多次買入賣出
思路：判斷第i天和第i-1天的價格差diff，把diff>0部分求和即可
計算相鄰股票價格的差值，計算差值大於0的元素的和，即爲最大利潤
class Solution(object):
def maxProfit(self, prices):
if prices == []: return 0
ans = 0
for i in range(1,len(prices)):
diff = prices[i]-prices[i-1]
if diff>0: ans+=diff
return ans- 714. Best Time to Buy and Sell Stock with Transaction Fee：買入賣出次數不限，但是每次需要支付一定的費用，求最大的收益,Medium
思路：
sell維護買入賣出價格(t[0],t[1])，其中t[1]爲0表示未賣出
如果棧頂未賣出，且price<=buy 替換買入
如果price大於max(賣出價格，買入價格+費用）替換賣出
price作爲買入價格入棧
當棧元素>1，並且合併棧頂的兩組交易可以獲得更大收益時，對棧頂的兩個交易進行合併
a2-a1-fee+b2-b1-fee< b2-a1-fee –> a2
class Solution(object):
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
sell = [[50000,0]]
for price in prices:
# 如果棧頂未賣出，且price<=buy 替換買入
if sell[-1][1]==0 and price<sell[-1][0]:
sell[-1][0] = price
# 如果price大於max(賣出價格，買入價格+費用）替換賣出
elif price>=max(sell[-1][0]+fee,sell[-1][1]):
sell[-1][1] = price
# price作爲買入價格入棧
elif sell[-1][1]:
sell.append([price,0])
# a2-a1-fee+b2-b1-fee<b2-a1-fee --> a2<b1+fee
while(len(sell)>1 and sell[-2][1]<sell[-1][0]+fee):
sell[-1][1] = max(sell.pop()[1],sell[-1][1])
return sum(t[1]-t[0]-fee for t in sell if t[1]-t[0]>fee)思路還是偏複雜了。。在kdd的指點，記兩個狀態，買p1和賣p2
p1 = max(p1,p2-price)
p2 = max(p2,p1+price-fee)
class Solution(object):
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
size = len(prices)
if size<=1: return 0
p1,p2 = -prices[0],0
for i in range(1,size):
p1 = max(p1,p2-prices[i])
p2 = max(p2,p1+prices[i]-fee)
return p2- 309. Best Time to Buy and Sell Stock with Cooldown：多次買入賣出，但是每次賣出後第二天不能買入，Medium
buy[i]表示時間爲i前操作爲買入的最大收益
sell[i]表示時間爲i前操作爲賣出的最大收益
rest[i]表示時間爲i前操作爲冷凍器的最大收益
有
buy[i] = max(buy[i-1],rest[i-1]-price) 前一天購買或者前一天冷凍當天購買
sell[i] = max(sell[i-1],buy[i-1]+price) 前一天賣出或者前一天購買當前賣出
rest[i] = max(rest[i-1],buy[i-1],sell[i-1])
buy[i]<=rest[i] 又有 rest[i] = sell[i-1] （因爲rest是在sell操作之後纔有的東西）
所以變成
buy[i] = max(buy[i-1],sell[i-2]-price)
sell[i] = max(sell[i-1],buy[i-1]+price)
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if len(prices)<2:
return 0
sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0
for price in prices:
prev_buy = buy
buy = max(prev_sell - price, prev_buy)
prev_sell = sell
sell = max(prev_buy + price, prev_sell)
return sell- 123. Best Time to Buy and Sell Stock III：最多買賣兩次，求最大利潤,Hard
計算相鄰股票價格的差值
遍歷i
prev[i]表示從左到右第i位的最大利潤
post[i]表示從右到左第i位的最大利潤
ans=max(ans,prev[i]+post[i])
class Solution(object):
def maxProfit(self,prices):
if prices == []: return 0
size = len(prices)
MIN,MAX = prices[0],prices[size-1]
prev,post = [0],[0]
for i in range(1,size):
x = prices[i]
y = prices[size-1-i]
MAX,MIN = max(MAX,y),min(MIN,x)
prev.append(max(prev[-1],x-MIN))
post.append(max(post[-1],MAX-y))
post.reverse()
ret = 0
for i in range(size):
ret = max(ret,prev[i]+post[i])
return retglobal G[i][j] 到第i天至多進行j次交易的最優利潤
local G[i][j] 到第i天至多進行j次交易的最優利潤，且在第i天賣出
L[i][j] = max(G[i-1][j-1],G[i-1][j-1]+diff,L[i-1][j]+diff)
說明：
G[i-1][j-1]爲第i-1天進行j-1次交易的最優利潤，在第i天沒有買入賣出行爲
G[i-1][j-1]+diff爲第i-1天進行j-1次交易，然後第i-1天買入，第i天賣出
L[i-1][j]+diff 爲第i-1天進行第j次交易，改爲第i天進行第j次交易
G[i][j] = max(G[i-1][j],L[i][j])
class Solution(object):
def maxProfit(self, k, prices):
"""
:type k: int
:type prices: List[int]
:rtype: int
"""
if prices == [] or k ==0 : return 0
size = len(prices)
if k>(size/2): return self.maxProfit1(prices)
L = [[0 for i in range(k+1)] for p in prices]
G = [[0 for i in range(k+1)] for p in prices]
for j in range(1,k+1):
for i in range(1,len(prices)):
diff = prices[i]-prices[i-1]
L[i][j] = max(G[i-1][j-1],G[i-1][j-1]+diff,L[i-1][j]+diff)
G[i][j] = max(G[i-1][j],L[i][j])
return G[len(prices)-1][k]
def maxProfit1(self, prices):
if prices == []: return 0
ans = 0
for i in range(1,len(prices)):
diff = prices[i]-prices[i-1]
if diff>0: ans+=diff
return ans
