标签：效率 正数 class col 就是 python实现 coin 时间复杂度 .so
【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)
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）
标签：效率 正数 class col 就是 python实现 coin 时间复杂度 .so
原文地址：https://www.cnblogs.com/ElfoDigger/p/10605620.html
