322. Coin Change

Posted 2021-02-12 ymjyqsx

 

https://www.cnblogs.com/grandyang/p/5138186.html

用动态规划的方法， dp[i]表示当前金额最少换零钱的次数

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