动态规划（5）

原题：

/** * Created by gouthamvidyapradhan on 22/03/2017. * You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin. * <p> * Note: You can assume that * <p> * 0 <= amount <= 5000 * 1 <= coin <= 5000 * the number of coins is less than 500 * the answer is guaranteed to fit into signed 32-bit integer * Example 1: * <p> * Input: amount = 5, coins = [1, 2, 5] * Output: 4 * Explanation: there are four ways to make up the amount: * 5=5 * 5=2+2+1 * 5=2+1+1+1 * 5=1+1+1+1+1 * Example 2: * <p> * Input: amount = 3, coins = [2] * Output: 0 * Explanation: the amount of 3 cannot be made up just with coins of 2. * Example 3: * <p> * Input: amount = 10, coins = [10] * Output: 1 */

答案：

public class CoinChange2 {    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        int[] coins = {1, 2, 5};        System.out.println(new CoinChange2().change(5, coins));    }    public int change(int amount, int[] coins) {        int[][] dp = new int[coins.length][amount + 1];        for (int i = 0, l = coins.length; i < l; i++)            Arrays.fill(dp[i], -1);        return dp(dp, 0, coins, coins.length, amount);    }    private int dp(int[][] dp, int i, int[] coins, int l, int n) {        if (n == 0) return 1;        else if (i >= l) return 0;        if (n < 0) return 0;        if (dp[i][n] != -1) return dp[i][n];        dp[i][n] = dp(dp, i + 1, coins, l, n) + dp(dp, i, coins, l, n - coins[i]);        return dp[i][n];    }}