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.
You may assume that you have an infinite number of each kind of coin.
Example1
Input:
[1, 2, 5]
11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example2
Input:
[2]
3
Output: -1
class Solution:
"""
@param coins: a list of integer
@param amount: a total amount of money amount
@return: the fewest number of coins that you need to make up
"""
def coinChange(self, coins, amount):
# dp[i][y] stands for the min number of coins needed, -1 if not possible
# first i coins considered
# amount y
dp = [[-1]*(amount+1) for i in range(len(coins)+1)]
dp[0][0] = 0
for i in range(1, len(coins)+1):
for y in range(amount+1):
if y < coins[i-1]:
dp[i][y] = dp[i-1][y]
else:
dp[i][y] = self.min_or(
dp[i-1][y],
self.add1(dp[i][y-coins[i-1]])
)
return dp[-1][-1]
def add1(self, a):
if a == -1:
return -1
return a + 1
def min_or(self, a, b):
if a == -1:
return b
if b == -1:
return a
return min([a, b])
-1
is very special key, so min and add operation are redefinedmin_or
里面,仍然符合交换律和结合律add1
函数里面