1、货币机找零问题的一般递归解法
代码
# @author:wpk # @time:2021 10 19 # @description: deal the problem change money with recursion import time def recMC(coinValueList,change): minCoins = change # 解决最小规模问题 if change in coinValueList: return 1 else: for i in [c for c in coinValueList if c <= change ]: # recMC(coinValueList,change - i)调用自身 + change - i减小规模 numCoins = 1 + recMC(coinValueList,change - i) if numCoins < minCoins: minCoins = numCoins return minCoins start = time.time() coinValueList =[1,5,10,25] # print(time.clock()) res = recMC([1,5,10,25],63) print("在零钱列表为%s的货币政策中,找零%s,需要的最小钱币个数为%s:" %(coinValueList,63,res)) end = time.time() print('usingtime = %s' %(end-start))
运行结果
非优化情况下的运行时间为: 26.27496314048767
2、优化后的递归解法
优化思路与理念
在解决找零过程中,递归解法存在多次的重复计算现象
解决方法:可以对计算过的钱币,进行记录,在下一次进行时候,先判断是否已经计算过,是,直接调用之前的结果;不是,再进行递归调用
思想:以储存空间,换计算时间
代码
# @author:wpk # @time:2021 10 19 # @description: deal the problem change money with recursion import time def recDC(coinValueList,change,KnownResults): minCoins = change # 递归基本结束条件 if change in coinValueList: #记录存储最优解 KnownResults[change] = 1 return 1 elif KnownResults[change] > 0: # 查表成功,直接调用计算过的最优解 return KnownResults[change] else: for i in [c for c in coinValueList if c <= change ]: numCoins = 1 + recDC(coinValueList,change - i,KnownResults) if numCoins < minCoins: minCoins = numCoins #找到的最优解,记录到列表中 KnownResults[change] = minCoins return minCoins start = time.time() coinValueList =[1,5,10,25] # print(time.clock()) res = recDC([1,5,10,25],63,[0]*64) print("在零钱列表为%s的货币政策中,找零%s,需要的最小钱币个数为%s:" %(coinValueList,63,res)) end = time.time() print('usingtime = %s' %(end-start))
优化效果
计算时间明显提升,几乎是瞬间得出!
3、补充:Python编程中计算程序运行时间的方法
import time start = time.time() run_function() end = time.time() print('usingtime = %s' %(end-start))
import time start = time.clock() run_function() end = time.clock() print('usingtime = %s' %(end-start))
import datetime start = datetime.datetime.now() run_function(): end = datetime.datetime.now() print('usingtime = %s' %(end-start))