题目
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.
Example 1:
Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins = [2], amount = 3
Output: -1
Example 3:
Input: coins = [1], amount = 0
Output: 0
Constraints:
1 <= coins.length <= 121 <= coins[i] <= 231 - 10 <= amount <= 104
思路
dp，开辟一个dp数组，索引表示amount，值表示最少所需钱的数量
代码
python版本：
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [0]+[99999]*amount
for i in range(1, len(dp)):
dp[i] = min([dp[i-j] for j in coins if i-j >= 0], default=99999)+1
return dp[-1] if dp[-1] < 99999 else -1
