难度: Medium
原题连接
内容描述
You are given coins of different denominations and a total amount of money amount. Write a function to compute 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.
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
Note:
You may assume that you have an infinite number of each kind of coin.
思路 1 - 时间复杂度: O(Namount)- 空间复杂度: O(amount)*****
DP入门
递推方程式: dp[i] = min(dp[i], dp[i-coins[j]]+1), coins[j] 是硬币的面额
class Solution(object):
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [amount+1] * (amount+1)
dp[0] = 0
for i in range(1, amount+1):
for coin in coins:
if coin <= i:
dp[i] = min(dp[i], dp[i-coin]+1)
return -1 if dp[-1] == amount + 1 else dp[-1]