斐波那契數(shù)列

斐波那契額數(shù)列：1,1,2,3,5,8,13,21,34
設(shè)計(jì)一個(gè)函數(shù)，求第n位上的數(shù)。

1.暴力解 很簡單的方法：

def fib(n):
    if n == 1 or n == 2:
        return 1
    else:
        return fib(n - 1) + fib(n - 2)

但是，這個(gè)方法有一個(gè)很大的問題，計(jì)算重復(fù)太多，例如，我們要算fib(5)

• fib(5)
  • fib(4)
    • fib(3)
      • fib(2)
      • fib(1)
    • fib(2)
  • fib(3)
    • fib(2)
    • fib(1)

2.DP algorithm

可以發(fā)現(xiàn)，當(dāng)中有大量的重復(fù)計(jì)算。我們可以設(shè)計(jì)一個(gè)備忘錄，將算過的數(shù)值保存進(jìn)去。

dic = {}
def fib2(n):
    if n in dic:
        return dic[n]
    if n == 0:
        return 0
    if n == 1:
        return 1
    value = fib2(n - 1) + fib2(n - 2)
    dic[n] = value
    return value

3.自下而上的DP algorithm

或者，從下而上進(jìn)行動(dòng)態(tài)規(guī)劃

dp = []
def fib3(n):
    dp = [0] * (n + 1)
    if n == 0:
        return 0
    dp[1] = 1
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    return dp[n]

比較fib(40),三種方法的時(shí)間分別是

27.43297505378723
3.62396240234375e-05
1.5020370483398438e-05

我們發(fā)現(xiàn)，時(shí)間成指數(shù)下降

我們發(fā)現(xiàn)，其實(shí)，f(n) 只和前兩個(gè)數(shù)字有關(guān)，所以，第三種方法沒有必要設(shè)置一個(gè)n+1長的盒子來保存數(shù)字，將fib3優(yōu)化：

def fib4(n):
    if n == 2 or n == 1:
        return 1
    prev = 1
    curr = 1
    for i in range(3, n + 1):
        sum = prev + curr
        prev = curr
        curr = sum
    return curr

時(shí)間進(jìn)一步降低：

6.9141387939453125e-06

找零錢

找零錢問題，給一個(gè)零錢數(shù)列:[1,2,5]，現(xiàn)在如果有11塊錢，問最少幾枚硬幣能湊出這個(gè)數(shù)額。

1.暴力解

如果要湊11，那么如果知道怎么湊10，就知道怎么湊11，因?yàn)橛?0，再加上一枚硬幣就可以得到11。

遞歸樹

但是需要選擇硬幣最少的那個(gè)結(jié)果：

def coinChange(amount):
    if amount == 0:
        return 0
    if amount < 0:
        return -1
    res = float('INF')
    for coin in coins:
        subproblem = coinChange(amount - coin)
        if subproblem == -1: continue
        # choose the min because we need the best choose, least coins --> minimum amount
        res = min(res, 1 + subproblem)
    return res if res != float('INF') else -1

2.DP algorithm

同上面斐波那契數(shù)列一樣，列一個(gè)備忘錄(memo)

memo = {}


def coinChange2(amount):
    if amount in memo:
        return memo[amount]
    if amount == 0:
        return 0
    if amount < 0:
        return -1
    res = float('INF')
    for coin in coins:
        subproblem = coinChange(amount - coin)
        if subproblem == -1: continue
        res = min(res, 1 + subproblem)
    memo[res] = res if res != float('INF') else -1
    return memo[res]

3.自下而上的DP algorithm

def coinChange3(amount):
    dp = [amount + 1] * (amount + 1)
    dp[0] = 0
    for item in range(len(dp)):
        for coin in coins:
            if (item - coin < 0): continue
            dp[item] = min(dp[item], 1 + dp[item - coin])
            print(dp)
    return -1 if (dp[amount] == (amount + 1)) else dp[amount]

可以發(fā)現(xiàn)，每一次的結(jié)果：

[0, 1, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 2, 12, 12, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 12, 12, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 12, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 12, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 3, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 12, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 3, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 3, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 12, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 12, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 3, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 12, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 12, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 4, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 12, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 4, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 4, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 12]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3]
[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3]

dynamic programming 就是一種聰明的窮舉