[LeetCode 322] Coin Change

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本文介绍了一个利用动态规划算法解决给定不同面额硬币组合成指定金额的最小数量问题的方法。通过实例演示了如何实现并优化算法以达到最优解。

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:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

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

Solution:

dp[amount] = Min (dp[diff] +1, dp[amount] )

public int coinChange(int[] coins, int amount) {
        if(coins == null || coins.length == 0) {
            return -1;
        }
        if(amount <=0) return 0;
        int[] dp = new int[amount+1];
        for(int i=1;i<dp.length; i++) {
            dp[i] = Integer.MAX_VALUE;
        }
        for(int am=1;am<=amount;am++) {
            for(int i=0;i<coins.length;i++) {
                if(coins[i]<=am) {
                    int diff = am - coins[i];
                    if(dp[diff] != Integer.MAX_VALUE) {
                        dp[am] = Math.min(dp[diff] +1, dp[am]);
                    }
                }   
            }
        }
        return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
    }