动态规划（4）

原题：

/** * Created by gouthamvidyapradhan on 23/06/2017. * 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. * <p> * Example 1: * coins = [1, 2, 5], amount = 11 * return 3 (11 = 5 + 5 + 1) * <p> * Example 2: * coins = [2], amount = 3 * return -1. * <p> * Note: * You may assume that you have an infinite number of each kind of coin. * <p> * Solution: * For example if you have N coins and amount equal to Q * For every coin you have two options * i) If you chose to include this coin then, total amount reduces by the sum equivalent to the value of this coin and you are * left with N coins and Q = (Q - value of this coin) * ii) If you chose not to include this coin then, you are left with N - 1 coins (since you chose to not to include this coin) * and total amount is still equal to Q * <p> * Calculate recursively for each coin and possible amount * Since there can be overlapping sub-problems you can save the state in a 2D matrix - a typical DP approach. * <p> * For each state minimum is calculated using -> Min(1 + fn(i, amount - v[i]), fn(i + 1, amount)) * <p> * Worst-case time complexity is O(N x Q) where N is the total number of coins and Q is the total amount */

答案：

public class CoinChange {    private int[][] DP;    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        int[] coins = {1, 2, 5};        System.out.println(new CoinChange().coinChange(coins, 11));    }    public int coinChange(int[] coins, int amount) {        DP = new int[coins.length][amount + 1];        int result = dp(amount, 0, coins);        if (result == Integer.MAX_VALUE - 1) return -1;        return result;    }    private int dp(int amount, int i, int[] coins) {        if (amount == 0) return 0;        else if (i >= coins.length || amount < 0) return Integer.MAX_VALUE - 1;        if (DP[i][amount] != 0) return DP[i][amount];        DP[i][amount] = Math.min(1 + dp(amount - coins[i], i, coins), dp(amount, i + 1, coins));        return DP[i][amount];    }}