如果我们有面值为1元、3元和5元的硬币若干枚，如何用最少的硬币凑够11元

def select_coin(coin_value, total_value):
min_coin_num = [0]
for i in range(1, total_value + 1):
min_coin_num.append(float(‘inf’))
for value in coin_value:
if value <= i:# and min_coin_num[i - value] + 1 < min_coin_num[i]:
min_coin_num[i] = min_coin_num[i - value] + 1

return min_coin_num

result = select_coin([1, 3, 5], 16)

动态规划思想 dp方程式如下

dp[0] = 0

dp[i] = min{dp[i - coins[j]] + 1}, 且 其中 i >= coins[j], 0 <= j < coins.length

回溯法，输出可找的硬币方案

path[i] 表示经过本次兑换后所剩下的面值，即 i - path[i] 可得到本次兑换的硬币值。

def changeCoins(coins, n):
if n < 0: return None
dp, path = [0] * (n+1), [0] * (n+1) # 初始化
for i in range(1, n+1):
minNum = i # 初始化当前硬币最优值
for c in coins: # 扫描一遍硬币列表，选择一个最优值
if i >= c and minNum > dp[i-c]+1:
minNum, path[i] = dp[i-c]+1, i - c
dp[i] = minNum # 更新当前硬币最优值
print(‘最少硬币数:’, dp[-1])
print(‘可找的硬币’, end=’: ‘)
while path[n] != 0:
print(n-path[n], end=’ ‘)
n = path[n]
print(n, end=’ ')

if name == ‘main’:
coins, n = [1, 3, 5], 22 # 输入可换的硬币种类，总金额n
changeCoins(coins, n)