题解
令dp[i][j]dp[i][j]dp[i][j]:使用coins[0...i]coins[0...i]coins[0...i]这么多的情况下,组成目标金额为j时的最少数量。coins[i]coins[i]coins[i]为第i个硬币
当不使用第i个硬币能取得目标金额获得的最少硬币数量为:
dp[i][j]=dp[i−1][j−0coins[i]]+0dp[i][j] = dp[i-1][j-0coins[i]]+0dp[i][j]=dp[i−1][j−0coins[i]]+0
当使用第i个硬币能取得目标金额,用最少的硬币数量为:
dp[i][j]=min(dp[i−1][j−k∗coins[i]]+k),k>=1且j−kcoins[i]≥0dp[i][j]=min(dp[i−1][j−k∗coins[i]]+k),k>=1且j-kcoins[i]≥0dp[i][j]=min(dp[i−1][j−k∗coins[i]]+k),k>=1且j−kcoins[i]≥0
故
dp[i][j]=min(dp[i−1][j],min(dp[i−1][j−k∗coins[i]]+k)),j−k∗coins[i]≥0dp[i][j] = min(dp[i-1][j],min(dp[i-1][j−k∗coins[i]]+k)),j-k*coins[i]≥0dp[i][j]=min(dp[i−1][j],min(dp[i−1][j−k∗coins[i]]+k)),j−k∗coins[i]≥0
dp[i][j]=min(dp[i−1][j],dp[i−1][j−cons[i]+1,dp[i−1][j−2cons[i]]+2,...,dp[i−1][j−kcons[i]]+k)(1)dp[i][j] = min(dp[i-1][j],dp[i-1][j-cons[i]+1,dp[i-1][j-2cons[i]]+2,...,dp[i-1][j-kcons[i]]+k) (1)dp[i][j]=min(dp[i−1][j],dp[i−1][j−cons[i]+1,dp[i−1][j−2cons[i]]+2,...,dp[i−1][j−kcons[i]]+k)(1)
令j=j-cons[i],将其带入上述公式得:
dp[i][j−cons[i]]=min(dp[i−1][j−cons[i],dp[i−1][j−2cons[i]]+1,...,dp[i−1][j−(k+1)cons[i]]+k)(2)dp[i][j-cons[i]] =min( dp[i-1][j-cons[i], dp[i-1][j-2cons[i]]+1,...,dp[i-1][j-(k+1)cons[i]]+k) (2)dp[i][j−cons[i]]=min(dp[i−1][j−cons[i],dp[i−1][j−2cons[i]]+1,...,dp[i−1][j−(k+1)cons[i]]+k)(2)
由于k=k+1,且将公式2加1得:
dp[i][j−cons[i]]+1=min(dp[i−1][j−cons[i]+1,dp[i−1][j−2cons[i]]+2,...,dp[i−1][j−(k+1)cons[i]]+k+1)(3)dp[i][j-cons[i]]+1 =min( dp[i-1][j-cons[i]+1, dp[i-1][j-2cons[i]]+2,...,dp[i-1][j-(k+1)cons[i]]+k+1) (3)dp[i][j−cons[i]]+1=min(dp[i−1][j−cons[i]+1,dp[i−1][j−2cons[i]]+2,...,dp[i−1][j−(k+1)cons[i]]+k+1)(3)
故公式(3)代入到公式(1)中得:
dp[i][j]=min(dp[i−1][j],dp[i][j−cons[i]]+1)dp[i][j] = min(dp[i-1][j],dp[i][j-cons[i]]+1)dp[i][j]=min(dp[i−1][j],dp[i][j−cons[i]]+1)
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
#利用动态规划求解
dp = [[amount + 1] * (amount + 1) for i in range(len(coins) + 1)]
#画网格进行求解
#初始化第一行为0
#进行初始化的设置]
for i in range(len(coins) + 1):
dp[i][0] = 0
#如果构不成目标金额,最后的值是amout + 1
for i in range(1,len(coins) + 1):
for j in range(1, amount + 1):
if j >= coins[i - 1]:
dp[i][j] = min(dp[i - 1][j], 1 + dp[i][j - coins[i - 1]])
else:
dp[i][j] = dp[i - 1][j]
if dp[-1][-1] == amount + 1:
return -1
else:
return dp[-1][-1]