算法之Python实现 - 002 ：换钱的最少货币数补充（每种货币只能使用一次） - Python

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【题目】：给定数组arr，arr中所有的值都为正数且不重复。每个值代表一种面值的货币，每种面值的货币仅可以使用一张，再给定一个整数aim代表要找的钱数，求组成aim的最少货币数。

【代码1】：时间与额外空间复杂度O(N*aim)

import numpy as np

from xmlrpc.client import MAXINT

def mincoin(arr,aim):

    if len(arr)<0:

        print("No coin provided for change!")

    if sum(arr)<aim:

        print("Coins provided not enough for change!")

    arr.sort()

    arr.reverse()

    if aim == 0:

        print("Aim is 0, no need to change!")

    i = 0

    j = 0

    maxval = 99#MAXINT

    dp = np.zeros((len(arr),aim+1))

    for i in range(1,len(arr)):

        dp[i] = np.array([99]*(aim+1))

        dp[i][0] = 0

    if arr[0]<= aim:

        dp[0][arr[0]] = 1

    left = 0

    for i in range(1,len(arr)):

        for j in range(1,aim+1):

            left = maxval

            if j-arr[i] >= 0 and dp[i-1][j-arr[i]] != maxval:

                left = dp[i-1][j-arr[i]]+1

            dp[i][j] = min(left,dp[i-1][j])

    #print(dp)

    #print('Need ',int(dp[aim]),' Coins.')

    print('Need ',int(dp[len(arr)-1][aim]),' Coins.')

# ===CALL === #

a = [5,2,3,5,8]

tar = 20

mincoin(a,tar)

【代码2】：时间复杂度O(N*aim)，额外空间复杂度O(aim)

import numpy as np

from xmlrpc.client import MAXINT

def mincoin(arr,aim):

    if len(arr)<0:

        print("No coin provided for change!")

    if sum(arr)<aim:

        print("Coins provided not enough for change!")

    arr.sort()

    arr.reverse()

    if aim == 0:

        print("Aim is 0, no need to change!")

    i = 0

    j = 0

    maxval = 99#MAXINT

    dp = np.array([99]*(aim+1))

    dp[0] = 0

    if arr[0]<= aim:

        dp[arr[0]] = 1

    left = 0

    for i in range(1,len(arr)):

        for j in range(1,aim+1):

            left = maxval

            if j-arr[i] >= 0 and dp[j-arr[i]] != maxval:

                left = dp[j-arr[i]]+1

            dp[j] = min(left,dp[j])

    #print(dp)

    #print('Need ',int(dp[aim]),' Coins.')

    print('Need ',int(dp[aim]),' Coins.')

# ===CALL === #

a = [5,2,3,5,8]

tar = 20

mincoin(a,tar)

【代码3】：时间复杂度O(N*aim)，额外空间复杂度O(aim)

同样的在原书也就是【代码2】的基础上，下面的执行效率会更高一点点，但是这种算法对于【代码1】的复杂度是有问题的。

import numpy as np

from xmlrpc.client import MAXINT

def mincoin(arr,aim):

    if len(arr)<0:

        print("No coin provided for change!")

    if sum(arr)<aim:

        print("Coins provided not enough for change!")

    arr.sort()

    arr.reverse()

    if aim == 0:

        print("Aim is 0, no need to change!")

    i = 0

    j = 0

    maxval = 99#MAXINT

    dp = np.array([99]*(aim+1))

    dp[0] = 0

    if arr[0]<= aim:

        dp[arr[0]] = 1

    left = 0

    for i in range(1,len(arr)):

        for j in range(j-arr[i],aim+1):

            left = maxval

            if dp[j-arr[i]] != maxval:

                left = dp[j-arr[i]]+1

            dp[j] = min(left,dp[j])

    #print(dp)

    #print('Need ',int(dp[aim]),' Coins.')

    print('Need ',int(dp[aim]),' Coins.')

# ===CALL === #

a = [5,2,3,5,8]

tar = 20

mincoin(a,tar)