518. Coin-Change-2

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

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

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

Method One

class Solution {
    // 这个题超级有意思。
    // 怪不得和这个 coin change 不一样
    // 相当于，我们先 dp 只用一种 coin;
    // 然后 再加上一种 coin 更新
    // 这样的话留下的就只是组合。
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for(int coin: coins ){
            for(int i = coin; i <= amount; i++){
                dp[i] += dp[i - coin];
            }
        }
        return dp[amount];
    }
}
​
​
/**
The fun part about this solution is that if you switch the order of the for loops in the code, your answer almost doubles! Learnt it the hard way.
​
To make an amount of 3 with two coin denominations 1 and 2, you can go:
1 + 2 = 3
2 + 1 = 3
Both mean the same thing, in this question.
​
所以下面的是错的。
​
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for(int i = 1; i <= amount; i++ ){
            for(int coin : coins ){
                if( i - coin < 0 ){
                    continue;
                }
                dp[i] += dp[i - coin];
            }       
        }
        return dp[amount];
    }
*/
​