Question

以下是针对最小硬币更换问题的强力解决方案。它需要一个int A，这是需要做出的改变，以及一系列硬币面额。它返回一个对象，结果，它具有可以根据硬币面额数组以及硬币数组返回的最小硬币。

例如，如果要求使用值10对[1, 2, 5]美分进行更改，则应返回2硬币分钟和数组[0, 0, 2]两分钱。< / p>

它返回正确的最小值但不是正确的硬币数组。

# values the algorithms should return, the min num of coins, the actual    
# coins in an array, and the original 

# array of coin denominations
class Results:

    a = 0
    change = []
    coinsDenom = []

# A is an array of coin denominations
# C is the change to be made
# returns results object
def changeslow(A, C):
    res = Results()
    res.coinsDenom = C
    res.change = []
    # initialize your change array to be the length of the coindenom array
    for i in range(0, len(res.coinsDenom)):
        res.change.append(0)

    minCoins = A

    for i in [j for j in C if j <= A]:
        if j == A:
            res.a = 1
            res.change[i] = res.change[i] + 1
            return res

        nextcall = changeslow(A-i, C)
        numCoins = 1 + nextcall.a

        if numCoins < minCoins:
            minCoins = numCoins
            res.change = nextcall.change
            res.change[0] = res.change[0] + 1



    res.a = minCoins

    return res

Answer 1

使用dynamic programming可以最好地解决此问题。例如，请参阅此http://www.columbia.edu/~cs2035/courses/csor4231.F07/dynamic.pdf以获取以下直观的动态编程公式：

# DP-CHANGE
# find number of coins needed to represent a value and memoize the last coin
def DP_CHANGE(denoms, value):
    num_coins = [0]*(value+1) # store number of coins needed to represent coin values from [0..value]
    last_coin = [float('Inf')]*(value+1) # memoize last coin used to represent coin values from [0..value]
    for d in denoms:
        num_coins[d] = 1
    for i in range(1, value + 1):
        num_denom = min([(num_coins[i-d] + 1, d) for d in denoms if i - d >= 0])
        num_coins[i], last_coin[i] = num_denom[0], num_denom[1]
    return num_coins, last_coin

# TRACE-CHANGE
# back-trace the denominations used
def TRACE_CHANGE(value, last_coin):
    denoms_used = []
    while value > 0:
        denoms_used.append(last_coin[value]);
        value-=last_coin[value]
    return denoms_used

def FIND_CHANGE(value, denoms):
    num_coins, last_coin = DP_CHANGE(denoms, value)
    print 'number of coins needed to represent values ' + str(range(value+1)) + ': ' + str(num_coins)
    print 'minimum number of coins need to represent the value ' + str(value) + ': ' + str(num_coins[value])
    print 'coins of denominations needed:' + str(TRACE_CHANGE(value, last_coin))

FIND_CHANGE(10, [1,2,5])
# number of coins needed to represent values [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]:
#                                            [0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2]
# minimum number of coins need to represent the value 10: 2
# coins of denominations needed:[5, 5]

分而治之 - 最小变化 - 将硬币作为数组返回

1 个答案: