斐波那契数列
斐波那契额数列：1,1,2,3,5,8,13,21,34
设计一个函数，求第n位上的数。
1.暴力解 很简单的方法：
def fib(n):
if n == 1 or n == 2:
return 1
else:
return fib(n - 1) + fib(n - 2)
但是，这个方法有一个很大的问题，计算重复太多，例如，我们要算fib(5)
- fib(5)
- fib(4)
- fib(3)
- fib(2)
- fib(1)
- fib(2)
- fib(3)
- fib(3)
- fib(2)
- fib(1)
- fib(4)
2.DP algorithm
可以发现，当中有大量的重复计算。我们可以设计一个备忘录，将算过的数值保存进去。
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，从下而上进行动态规划
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]
比较fib(40),三种方法的时间分别是
27.43297505378723
3.62396240234375e-05
1.5020370483398438e-05
我们发现，时间成指数下降
我们发现，其实，f(n) 只和前两个数字有关，所以，第三种方法没有必要设置一个n+1长的盒子来保存数字，将fib3优化：
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
时间进一步降低：
6.9141387939453125e-06
找零钱
找零钱问题，给一个零钱数列:[1,2,5]，现在如果有11块钱，问最少几枚硬币能凑出这个数额。
1.暴力解
如果要凑11，那么如果知道怎么凑10，就知道怎么凑11，因为有10，再加上一枚硬币就可以得到11。
递归树
但是需要选择硬币最少的那个结果：
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
同上面斐波那契数列一样，列一个备忘录(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]
可以发现，每一次的结果：
[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 就是一种聪明的穷举
