LeetCode 518. Coin Change 2

[woodyiiiiiii]

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

---

首先，我一开始想到的是根据Coin change 1的思路，dp[j] += dp[j - coins[i - 1]]，然而会出现重复的情况 ，比如coins=[1, 2, 5]，当amount == 3时，会出现3 = 1 + 2, 3 = 2 + 1的情况。如何排除多余的情况呢，一开始我想利用最后结果/=2来排除，但会超出整数范围。
既然这样，我们细分dp的状态，将一维数组扩展成二维数组，dp[i][j]表示在j钱下使用了只使用coins[0] ~ coins[i - 1]钱币的组合方法数。为了不让上述重复情况和遗漏情况的发生，我们只算当前coins[i - 1]的方法数，一个一个硬币相加，而不去考虑i - 1之前的硬币使用。比如coins = [1, 2]，amount = 3时，当i = 1，我们只算硬币2的增加，所以只会出现[1, 2]，不会出现[2, 1]。

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

当然也可以优化成一维数组：

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

参考资料2有三种递归解法，可以去看一下。

---

参考资料:

1. LeetCode
2. grandyang/leetcode#518