322. 零钱兑换

class Solution 
{
public:
    int coinChange(vector<int>& coins, int amount) 
    {
        vector<long long> dp(amount + 1,INT_MAX);
        dp[0] = 0;//总金额为0元所需的最少的硬币个数为0个
        for(int i = 0;i < coins.size();i ++)
        {
            for(int j = coins[i];j <= amount;j ++)
            {
                dp[j] = min(dp[j],dp[j - coins[i]] + 1);
            }
        }
        return dp[amount] == INT_MAX ? -1 : dp[amount];
    }
};