題目鏈接

tag:

• Medium;
• Dynamic Programming;

question:
??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

思路：
??建立dp數組，保存能到達當前amount的步數。逐個金額遍歷，看只用前j個金額能到達i的步數有多少，遞推式為：dp[i] = min(dp[i], dp[i-coins[j]] + 1);。代碼如下：

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 (coins[j] <= i) {
                    dp[i] = min(dp[i], dp[i-coins[j]] + 1); 
                }
            }
        }
        return (dp[amount] > amount) ? -1 : dp[amount];
    }
};