1. 题目
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。
你可以认为每种硬币的数量是无限的。
示例 1：
输入：coins = [1, 2, 5], amount = 11
输出：3
解释：11 = 5 + 5 + 1
示例 2：
输入：coins = [2], amount = 3
输出：-1
示例 3：
输入：coins = [1], amount = 0
输出：0
示例 4：
输入：coins = [1], amount = 1
输出：1
示例 5：
输入：coins = [1], amount = 2
输出：2
提示：
1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104
2. 代码
2.1 解法1
import math
from typing import List
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [math.inf] * (amount + 1)
dp[0] = 0
for i in range(1, amount + 1):
for coin in coins:
diff = i - coin # 还剩多少
if diff >= 0:
dp[i] = min(dp[i], dp[diff] + 1)
return dp[amount] if dp[amount] != math.inf else -1
2.2 解法2
import math
from typing import List
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
# 初始化数组db
dp = [-1] * (amount + 1)
dp[0] = 0
for i in range(amount + 1): # i 表示总金额
for coin in coins:
diff = i - coin
if diff >= 0 and dp[diff] != -1:
if dp[i] == -1 or dp[i] > dp[diff] + 1:
dp[i] = dp[diff] + 1
return dp[amount]
2.3 解法3
import math
from typing import List
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [0] + [10001] * amount
for coin in coins:
for j in range(coin, amount + 1):
dp[j] = min(dp[j], dp[j - coin] + 1)
return dp[amount] if dp[amount] != 10001 else -1
