【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):
python算法题,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!")
量化 python。dp = np.zeros((len(arr),aim+1))
i = 0
j = 0
left = aim
maxval = MAXINT
for j in range(1,aim+1):
python的int,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
python算法详解?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]
python 排序?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):
python币,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!")
python 类,dp = np.zeros((1,aim+1))[0]
i = 0
j = 0
maxval = MAXINT
for j in range(1,aim+1):
dp[j] = maxval
python数据结构,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
数字货币跨期套利python,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 === #
用python进行货币转换、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
python货币转换程序代码,from xmlrpc.client import MAXINT
def mincoin(arr,aim):
if len(arr)<0:
print("No coin provided for change!")
arr.sort()
arr.reverse()
python算法大全,if aim == 0:
print("Aim is 0, no need to change!")
dp = np.zeros((1,aim+1))[0]
i = 0
j = 0
maxval = MAXINT
python外汇交易,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)
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