518 Coin Change 2

518. Coin Change 2

1. Question

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

2. Implementation

(1) 完全背包

思路: 注意这里要求的是组合而不是排列, 所以第一层的for loop是coins数组中每个可能的coin，而第二层for loop是amount。如果要求排列的话，两个for loop的顺序要反过来。dp[i]表示对于amount为i，需要的coins的combination

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

3. Time & Space Complexity

完全背包: 时间复杂度O(n * amount), 空间复杂度O(amount)