## 题目描述：

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:
coins = `[1, 2, 5]`, amount = `11`
return `3` (11 = 5 + 5 + 1)

Example 2:
coins = `[2]`, amount = `3`
return `-1`.

Note:
You may assume that you have an infinite number of each kind of coin.

## 解题思路：

`dp[x + c] = min(dp[x] + 1, dp[x + c])`

## Python代码：

``````class Solution(object):
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [0] + [-1] * amount
for x in range(amount):
if dp[x] < 0:
continue
for c in coins:
if x + c > amount:
continue
if dp[x + c] < 0 or dp[x + c] > dp[x] + 1:
dp[x + c] = dp[x] + 1
return dp[amount]
``````

## Python代码：

``````class Solution(object):
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
steps = collections.defaultdict(int)
queue = collections.deque([0])
steps[0] = 0
while queue:
front = queue.popleft()
level = steps[front]
if front == amount:
return level
for c in coins:
if front + c > amount:
continue
if front + c not in steps:
queue += front + c,
steps[front + c] = level + 1
return -1
``````

Pingbacks已关闭。