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.

Note: You can assume that

• 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:

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:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10] 
Output: 1

法一超時

public class Solution {
    public int change(int amount, int[] coins) {
        Arrays.sort(coins);
        if(coins.length==0){
            if(amount==0){
                return 1;
            }
            return 0;
        }
        int temp=coins[coins.length-1];
        help(amount,coins,temp);
        return count;
    }
    int count=0;
    private void help(int amount, int[] coins,int mark){
        if(amount==0){
            count++;
            return;
        }else if(amount<0){
            return;
        }
        for(int i=0;i<coins.length;i++){
            if(mark>=coins[i]){
                int temp=coins[i];
                int tempamount=amount-temp;
                help( tempamount,  coins, temp);
            }
        }
    }
}
 

方法二採用了動態規劃。。想了半天，沒想出來。。看得別人的狀態轉移方程

public class Solution {
    public int change(int amount, int[] coins) {
        int[] table=new int[amount+1];
        table[0]=1;
        for(int i=0;i<coins.length;i++){
            int temp=coins[i];
            for(int j=0;j<amount+1;j++){
                if(j-coins[i]>=0){
                    table[j]+=table[j-coins[i]];
                }
            }
        }
        return table[amount];
    }
}