# 322. Coin Change 零钱兑换

@TOC

## # 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:

``````Input: coins = [1, 2, 5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
``````

Example 2:

``````Input: coins = [2], amount = 3
Output: -1
``````

Note:

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

## # 解题方法

### # 动态规划

``````class Solution(object):
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [float('inf')] * (amount + 1)
dp[0] = 0
for coin in coins:
for i in range(coin, amount + 1):
if dp[i - coin] != float('inf'):
dp[i] = min(dp[i], dp[i - coin] + 1)
return -1 if dp[amount] == float('inf') else dp[amount]
``````

C++代码如下：

``````class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
const int N = coins.size();
vector<int> dp(amount + 1, INT_MAX);
dp[0] = 0;
for (int coin : coins) {
for (int i = coin; i <= amount; ++i) {
if (dp[i - coin] != INT_MAX) {
dp[i] = min(dp[i], dp[i - coin] + 1);
}
}
}
return dp[amount] == INT_MAX ? -1 : dp[amount];
}
};
``````