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  • 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)   
    

      

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  • 原文地址:https://www.cnblogs.com/ElfoDigger/p/10600706.html
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