3递归（动态规划思想）

Posted 2021-02-20 aluosen

# -*- coding:utf-8 -*-
#@Time : 2020/5/2 11:06
#@Author: Aluosen
#@File : Recurrence.py
##进制转换
def toStr(n,base):
    convertString = ‘0123456789ABCDEF‘
    if n < base:
        return convertString[n]
    else:
        return toStr(n//base,base) + convertString[n%base]
print(toStr(65536,16))

#汉诺塔问题
def moveTower(height, fromPole, withPole, toPole):
    if height >= 1:
        moveTower(height - 1, fromPole, toPole, withPole)
        moveDisk(height, fromPole, toPole)
        moveTower(height - 1, withPole, fromPole, toPole)

def moveDisk(disk, fromPole, toPole):
    print(f‘Movin disd[{disk}] from {fromPole} to {toPole}‘)
moveTower(3,‘#1‘,‘#2‘,‘#3‘)

 

 

#零钱兑换问题-1、递归解法
def recMC(coinValueList,change):
    minCoins = change
    if change in coinValueList:                             #最小规模
        return 1
    else:
        for i in [c for c in coinValueList if c <= change]: #调用自身
            numCoins = 1 + recMC(coinValueList, change - i) #减小规模
            if numCoins < minCoins:
                minCoins = numCoins
    return minCoins

#零钱兑换问题-2、递归解法改进（增加了最优解表,中间结果记录）
def recMC(coinValueList,change,knownResults):
    minCoins = change
    if change in coinValueList:                             #最小规模，递归基本结束
        knownResults[change] = 1                            #记录最优解
        return 1
    elif knownResults[change] > 0:
        return knownResults[change]                         #查表成功
    else:
        for i in [c for c in coinValueList if c <= change]: #调用自身
            numCoins = 1 + recMC(coinValueList, change - i,knownResults) #减小规模
            if numCoins < minCoins:
                minCoins = numCoins
                #找到最优解，记录到表中
                knownResults[change] = minCoins
    return minCoins

#零钱兑换问题-3、动态规划算法
def dpMakeChange(coinValueList, change, minCoins):
    for cents in range(1, change + 1):
        coinCount = cents
        for j in [c for c in coinValueList if c <= cents]:
            if minCoins[cents -j] + 1 < coinCount:
                coinCount = minCoins[cents -j] +1
        minCoins[cents] = coinCount
    return minCoins[change]