问题描述
我正在练习动态编程的概念（递归不是我的强项）。我想知道如何改进我的代码，以便避免堆栈溢出。
任何帮助，谢谢！
def coinFlipping(n):
"""
For n amount of change,return the minimal amount of currency in coins.
Top-Down Approach (Memoization): Memo dictionary stores already solved
subproblems to reuse and avoid recomputing. It's essentially recursion
but optimized to avoid recomputing results.
"""
COINS = {
"penny": 0.01,"nickel": 0.05,"dime": 0.10,"quarter": 0.25,"dollar": 1.00
}
memo = {}
def __coinFlipping(n):
if n not in memo:
# Base case (if n is equal to 0.01,0.05,0.10,0.25..)
if n in COINS.values():
memo[n] = 1
else:
results = []
coins = (coin for coin in COINS.values() if coin < n)
for coin in coins:
results.append(__coinFlipping(n - coin))
memo[n] = 1 + min(results)
else:
return memo[n]
__coinFlipping(n)
print(memo[n])
coinFlipping(10)
我获得了以下错误：
File "/mnt/d/programs/algorithms/dynamicProgramming.py",line 45,in __coinFlipping
results.append(__coinFlipping(n - coin))
File "/mnt/d/programs/algorithms/dynamicProgramming.py",in __coinFlipping
results.append(__coinFlipping(n - coin))
[PrevIoUs line repeated 993 more times]
File "/mnt/d/programs/algorithms/dynamicProgramming.py",line 44,in __coinFlipping
for coin in coins:
File "/mnt/d/programs/algorithms/dynamicProgramming.py",line 43,in <genexpr>
coins = (coin for coin in COINS.values() if coin < n)
RecursionError: maximum recursion depth exceeded in comparison
解决方法
我将把我的最终代码留给以后查看该帖子的任何人。
def coinFlipping(n):
"""
For n amount of cents,return the minimal amount of currency in coins.
Top-Down Approach (Memoization): Memo dictionary stores already solved
subproblems to reuse and avoid recomputing. It's essentially recursion
but optimized to avoid recomputing results.
"""
COINS = {
"dollar": 100,"quarter": 25,"dime": 10,"nickel": 5,"penny": 1
}
memo = {}
def __coinFlipping(change):
if change not in memo:
# Base case (if n is equal to 0.01,0.05,0.10,0.25..)
if change in COINS.values():
memo[change] = 1
else:
results = []
coins = [coin for coin in COINS.values() if coin < change]
for coin in coins:
results.append(__coinFlipping(change - coin))
memo[change] = 1 + min(results)
return memo[change]
__coinFlipping(n)
print(memo[n])
coinFlipping(42)
实现此示例后，我开始认为这是一个不好的记忆示例。我本可以编写一个贪心算法来选择每一步中的最大价值硬币。例如从42c减去25c，从17c减去10c，依此类推，同时增加一个计数器。
,user2864740发现了。更改为整数，代码开始运行。