一 用python实现一次买入和卖出
1.时间复杂度为o(n^2)的方法
# import matplotlib as mpl
# import matplotlib.pyplot as plt
# pip install matplotlib
import mathfrom BuySellPair
import BuySellPairfrom Point
import Pointy = []
for x in range(0, 60): y.append(math.sin(math.pow(x - 35, 2) / 100) * (x - 35))
# plt.figure(1)
# plt.plot(x, y)
# plt.show()
# 最大值记为max
benefit = 0buy = 0sell = 0
# 先在函数内 遍历每一个点（第一层循环）作为买入点
# 找到该点之后，任意一个点（卖出店）与该点的差值（盈利），并与max比较，比max大则标记下来
for i in range(0, len(y)): before = y[i] for j in range(i, len(y)): after = y[j] dy = after - before if benefit < dy: benefit = dy buy = i sell = jprint("买入点为%s，卖出点位%s，获利：%s" % (buy, sell, benefit))
2.时间复杂度为O(n)的方法
pare = BuySellPair()
pare.benefit = 0
minPricePoint = Point()
for i in range(1, len(y)): benefit_temp = y[i] - pare.buy.y
if benefit_temp > pare.benefit:
pare.sell = Point(i, y[i])
pare.benefit = benefit_temp
if y[i] < minPricePoint.y:
minPricePoint = Point(i, y[i])
possible_max = y[i] - minPricePoint.y
if possible_max > pare.benefit: pare.sell = Point(i, y[i]) pare.benefit = possible_max pare.buy = minPricePointprint(pare)
二.用js实现
1.时间复杂度为O(n)的方法
