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

方法1：(動態規劃)

package leetcode;

import org.junit.Test;

public class CoinChange {
	@Test
	public void fun() {
		int coins[] = { 1, 2, 5 };
		int amount = 11;
		int method = CoinChange(coins, amount);
		System.out.println(method);
	}

	private int CoinChange(int[] coins, int amount) {
		// TODO Auto-generated method stub
		if (amount == 0) {
			return 0;
		}
		if (coins == null || coins.length == 0) {
			return -1;
		}
		int[] dp = new int[amount + 1];
		for (int i = 1; i <= amount; i++) {
			int res = Integer.MAX_VALUE;
			for (int j = 0; j < coins.length; j++) {
				if (coins[j] > i) {
					continue;
				}
				if (dp[i - coins[j]] == -1) {
					continue;
				}
				dp[i] = 1 + dp[i - coins[j]];
				res = Math.min(res, dp[i]);
			}
			dp[i] = res;
		}
		return dp[amount];
	}
}
 

時間複雜度：O(n^2)

空間複雜度：O(n)