python整钱兑换零钱基础写法_算法之Python实现 - 001 ： 换钱的最少货币数
【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!")
arr.sort()
arr.reverse()
if aim == 0:
print("Aim is 0, no need to change!")
dp = np.zeros((len(arr),aim+1))
i = 0
j = 0
left = aim
maxval = MAXINT
for j in range(1,aim+1):
dp[0][j] = maxval
if j-arr[0] >=0 and dp[0][j-arr[0]] != maxval:
dp[0][j] = dp[0][j-arr[0]]+1
for i in range(1,len(arr)):
for j in range(1,aim+1):
left = maxval
if j-arr[i] >=0 and dp[i][j-arr[i]] != maxval:
left = dp[i][j-arr[i]]+1
dp[i][j] = min(left,dp[i-1][j])
print('Need ',int(dp[len(arr)-1][aim]),' Coins.')
===CALL ===
a = [3,5,2]
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!")
arr.sort()
arr.reverse()
if aim == 0:
print("Aim is 0, no need to change!")
dp = np.zeros((1,aim+1))[0]
i = 0
j = 0
maxval = MAXINT
for j in range(1,aim+1):
dp[j] = maxval
if j-arr[0] >=0 and dp[j-arr[0]] != maxval:
dp[j] = dp[j-arr[0]]+1
left = 0
for i in range(1,len(arr)-1):
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.')
===CALL ===
a = [5,2,3]
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)
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!")
arr.sort()
arr.reverse()
if aim == 0:
print("Aim is 0, no need to change!")
dp = np.zeros((1,aim+1))[0]
i = 0
j = 0
maxval = MAXINT
for j in range(1,aim+1):
dp[j] = maxval
if j-arr[0] >=0 and dp[j-arr[0]] != maxval:
dp[j] = dp[j-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.')
===CALL ===
a = [5,2,3]
tar = 20
mincoin(a,tar)