[LeetCode] 由 「找零錢" 所想

 

Ref: [Optimization] Dynamic programming【尋找子問題】

 

Ref: [Optimization] Advanced Dynamic programming【優於recursion】

 

 

 

顯然，本篇是關於」動態規劃"的部分。

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找零錢 

1、簡單直白策略

要點：」個數「其實表明了「for循環」的層數。但「個數」不定，使用「遞歸」反而能解決這個問題，減小思路上的負擔。 

import time 

def recMC(coinValueList, change):

    minCoins = change
    if change in coinValueList:
        return 1
    else:
        for i in [c for c in coinValueList if c <= change]:
            # 遍歷每個變量, 至關於多重循環。有意思的是，循環的深度是不肯定的;
            numCoins = 1 + recMC(coinValueList, change-i)

            # 獲得這個遍歷分支的最終結果；
            if numCoins < minCoins:
                minCoins = numCoins

    return minCoins


start = time.time()
print(recMC([1,5,10,25], 63))
end = time.time()
print(end-start)

耗時：31秒。

 

 

2、記憶化策略

要點：增長了」結果緩存「，效率極具增長。

 

代碼中的 knownResults 記錄了上圖中節點（某個change時）的最優/最小的值，做爲了記錄。

def recDC(coinValueList, change, knownResults):

    minCoins = change
    if change in coinValueList:
        knownResults[change] = 1　　# 一個硬幣剛恰好
        return 1
    elif knownResults[change] > 0: return knownResults[change]     else: 
        for i in [c for c in coinValueList if c <= change]:
            # 
            numCoins = 1 + recDC(coinValueList, change-i, knownResults)

            # 更新minCoins
            if numCoins < minCoins: 
                minCoins = numCoins
                knownResults[change] = minCoins     return minCoins


def main():
    amnt = 2163
    print(recDC([1,2,5,10,21,25], amnt, [0]*(amnt+1)))

main()

耗時：0.01秒。

 

 

3、動態規劃策略

　　要點：不是從上到下的 (遞歸) 思惟，且容易棧溢出；而是從最小的狀況入手，逐漸擴大。

def dpMakeChange(coinValueList, change, minCoins, coinsUsed):
    # 首先，考慮全部的 "狀況"，是從小到大考慮
    for cents in range(change+1):

        coinCount = cents
        newCoin = 1

        # 考慮這些 "狀況" 下的可行的 "硬幣"
        for j in [c for c in coinValueList if c <= cents]:
            # 拿掉這枚"硬幣"，考慮 "上一個問題"的最優值
            if minCoins[cents-j] + 1 < coinCount: 
                coinCount = minCoins[cents-j] + 1
                newCoin = j
        
        # 考慮過該 "狀況"，更新記錄
        minCoins[cents]  = coinCount
        coinsUsed[cents] = newCoin
    
    return minCoins[change]


def printCoins(coinsUsed, change):
    coin = change
    while coin > 0:
        thisCoin = coinsUsed[coin]
        print(thisCoin)
        coin = coin - thisCoin

@fn_timer
def main():
    amnt = 2163
    clist = [1,2,5,10,21,25]
    coinsUsed = [0]*(amnt+1)
    coinsCount = [0]*(amnt+1)

    # print("Making change for", amnt, "requires")
    print(dpMakeChange(clist, amnt, coinsCount, coinsUsed), "coins")
    # print("They are:")
    # printCoins(coinsUsed,amnt)
    # print("The used list is as follows:")
    # print(coinsUsed)

main()

耗時：0.003秒。

 

End.