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）

Posted 2021-11-10 elfodigger

【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)