# 给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额,返回
# -1。
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# 你可以认为每种硬币的数量是无限的。
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# 示例 1:
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# 输入:coins = [1, 2, 5], amount = 11
# 输出:3
# 解释:11 = 5 + 5 + 1
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# 示例 2:
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# 输入:coins = [2], amount = 3
# 输出:-1
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# 示例 3:
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# 输入:coins = [1], amount = 0
# 输出:0
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# 示例 4:
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# 输入:coins = [1], amount = 1
# 输出:1
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# 示例 5:
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# 输入:coins = [1], amount = 2
# 输出:2
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# 提示:
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# 1 <= coins.length <= 12
# 1 <= coins[i] <= 231 - 1
# 0 <= amount <= 104
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# Related Topics 动态规划
# 👍 967 👎 0
# leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
def __init__(self):
self.cache = dict()
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
# base case
if amount == 0:
return 0
if amount < 0:
return -1
# 缓存
if amount in self.cache:
return self.cache[amount]
res = float("inf")
for coin in coins:
pre = self.coinChange(coins, amount - coin)
self.cache[amount - coin] = pre
if pre == -1:
continue
res = min(res, 1 + pre)
return -1 if res == float("inf") else res
# leetcode submit region end(Prohibit modification and deletion)
print(Solution().coinChange([11, 12], 25))
自底向上(推荐)
# leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
def __init__(self):
self.cache = dict()
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [float("inf") for x in range(amount + 1)]
dp[0] = 0
for i in range(amount + 1):
for coin in coins:
if i < coin:
continue
dp[i] = min(dp[i], 1 + dp[i - coin])
return dp[amount] if dp[amount] != float("inf") else -1
# leetcode submit region end(Prohibit modification and deletion)
print(Solution().coinChange([11, 12], 2000002))