算法：找零錢322. Coin Change

# 322. Coin Change

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return 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.

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

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
           

Example 3:

Input: coins = [1], amount = 0
Output: 0
           

Example 4:

Input: coins = [1], amount = 1
Output: 1
           

Example 5:

Input: coins = [1], amount = 2
Output: 2
           

Constraints:

1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104
           

動态規劃解法

動态規劃轉換公式

dp[i] = Math.min(dp[i], dp[i - coin] + 1);

。

這裡的要點是，如果有值，那麼一定要比預設值小。并且要周遊完所有的值。

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin: coins) {
                if (i - coin >= 0) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        
        return dp[amount] != amount + 1 ? dp[amount] : -1;
    }
}