算法之Python实现

【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，每种面值的货币可以使用任意张，再给定一个整数aim代表要找的钱数，求组成aim的方法数。

【代码1】递归

import numpy as np

def changemeans(arr,aim):
    if len(arr)<0:
        print("No coin provided for change!")
    arr.sort()
    arr.reverse()
    m = process(arr,0,aim)
    print('There are ',m,' ways!')

def process(arr,idx,aim):
    res = 0
    i = 0
    if aim == 0:
        res = 1
    else :
        if idx == len(arr):
            res = 0
        else :
            while arr[idx]*i <= aim:
                res += process(arr,idx+1,aim - arr[idx]*i)
                i += 1
    return res
        
    
# ===CALL === #
a = [5,10,25,1]
tar = 1000
changemeans(a,tar)    

【代码2】改进递归（递归加入记忆搜索）：时间复杂度O(N * aim²)

　　【原理】：例如按照题目中的a = [5,10,25,1]，使用a[0]和a[1]，利用[25,1]组成剩余的980元的可能性就是一种重复递归，假设利用[25,1]组成剩余的980元需要5秒钟，那么【代码1】需要搜索5*0+10*2,5*2+10*1,5*5 三次递归，【代码2】额外耗用了O((N+1)*(aim+1))的空间，但是只要三次寻址即可。

import numpy as np

def changemeans(arr,aim):
    if len(arr)<0:
        print("No coin provided for change!")
    arr.sort()
    arr.reverse()
    map = np.zeros((len(arr)+1,aim+1))
    m = process(arr,0,aim,map)
    print('There are ',m,' ways!')

def process(arr,idx,aim,map):
    res = 0
    i = 0
    if aim == 0:
        res = 1
    else :
        if idx == len(arr):
            res = 0
        else :     
            while arr[idx]*i <= aim:
                mapval = map[idx+1][aim- arr[idx]*i]
                if mapval != 0:
                    if mapval == -1:    mapval = 0
                    res += mapval
                else:   
                    res += process(arr,idx+1,aim - arr[idx]*i,map)
                i += 1
    if res == 0:
        map[idx][aim] = -1
    else :
        map[idx][aim] = res
    #print(':',int(map[idx][aim]),res)
    return res
        
    
# ===CALL === #
a = [5,10,25,1]
tar = 1000
changemeans(a,tar)    

 【代码3】：时间复杂度O(N * aim²)

import numpy as np

def changemeans(arr,aim):
    n = len(arr)
    if n<=0:
        print('No coin provided for exchange.')
    j = 0
    dp = np.zeros((n,aim+1))
    
    for i in range(0,n):
        dp[i][0] = 1
    
    while j*arr[0]<= aim:
        dp[0][j*arr[0]] = 1
        j += 1
    
    for i in range(1,n):
        for j in range(1,aim+1):
            num  = 0
            k = 0
            while j-arr[i]*k >= 0:
                num += dp[i-1][j-arr[i]*k]
                k += 1
            dp[i][j] = num
    
    print(dp[n-1][aim])
    
# ===CALL === #
a = [5,10,25,1]
tar = 1000
changemeans(a,tar)      

【代码4】：

另外实际上算arr[0..i-1]的组成剩下的方法，只会从最少的那个钱币为下标的位置开始，因此可以改为：

import numpy as np

def changemeans(arr,aim):
    n = len(arr)
    if n<=0:
        print('No coin provided for exchange.')
    j = 0
    dp = np.zeros((n,aim+1))
    
    for i in range(0,n):
        dp[i][0] = 1
    
    while j*arr[0]<= aim:
        dp[0][j*arr[0]] = 1
        j += 1
    
    for i in range(1,n):
        for j in range(min(arr)-1,aim+1):
            num  = 0
            k = 0
            while j-arr[i]*k >= 0:
                num += dp[i-1][j-arr[i]*k]
                k += 1
            dp[i][j] = num
    
    print(dp[n-1][aim])
    
# ===CALL === #
a = [5,10,25,2]
tar = 1000
changemeans(a,tar)