dp[i]代表要湊齊總數為i的錢所需要的最少硬幣數
dp[i]=min(dp[i]-coins[j])+1 for all j

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int min=0x7fffffff;
        vector<int> dp(amount+1,0);
        dp[0]=0;
        dp[1]=1;
        for(int i=1;i<=amount;i++){
            for
 
(int j=0;j<coins.size();j++){
                if(i-coins[j]>=0&&dp[i-coins[j]]<amount+1){
                    min=dp[i-coins[j]];
                }
            }
            dp[i]=min+1;
            min=0x7fffffff;
        }
        return dp[amount];
    }
};

這裡面考慮到有面值為1的硬幣的情況
測試用例

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int min=0x7fffffff;
        vector<int> dp(amount+1,amount+1);  //amount+1為不可能需要的次數，湊不出時作為判斷條件
        dp[0]=0;
       // dp[1]=1;
        for(int i=1;i<=amount;i++){
            for
 
(int j=0;j<coins.size();j++){
                if(i-coins[j]>=0&&dp[i-coins[j]]<amount+1){ //如果i-coins[j]能夠湊出來
                    if(dp[i-coins[j]]<min)  min=dp[i-coins[j]];  
                }
            }
            if(min!=0x7fffffff)  dp[i]=min+1;  //如果能夠湊出來
            min=0x7fffffff;
        }
        return dp[amount]==amount+1?-1:dp[amount];  
    }
};