[LeetCode] 518. Coin Change 2 硬币找零 2

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

给定一些不同面值的硬币，和一个钱数。编写函数计算要得到目标金额有多少种不同的硬币组合方式。

322. Coin Change 的变形，322题是求最少能用几个硬币组成给的钱数，而这题求的是组成给定钱数总共有多少种不同的方法。

解法：动态规划DP, 建立dp数组，保存能到达当前amount的步数。逐个金额遍历，看只用前i个金额能到达j的步数有多少，实现方法是累加起来dp[当前amount - 第i个金额]，最后返回dp[amount]。

State: dp[i], 表示总额为i时的方案数

Function: dp[i] = Σdp[i - coins[j]],  表示总额为i时的方案数 = 总额为i-coins[j]的方案数的加和

Initialize: dp[0] = 1, 表示总额为0时方案数为1

Retrun: dp[n] or dp[-1]

Java:

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

Python:

class Solution(object):
    def change(self, amount, coins):
        """
        :type amount: int
        :type coins: List[int]
        :rtype: int
        """
        dp = [0] * (amount + 1)
        dp[0] = 1
        for c in coins:
            for x in range(c, amount + 1):
                dp[x] += dp[x - c]
        return dp[amount]　

扩展思考：将上述代码中的循环顺序对调，即为求不同硬币的排列数（Permutation）

class Solution(object):
    def change(self, amount, coins):
        """
        :type amount: int
        :type coins: List[int]
        :rtype: int
        """
        dp = [0] * (amount + 1)
        dp[0] = 1
        for x in range(amount + 1):
            for c in coins:
                if c > x: continue
                dp[x] += dp[x - c]
        return dp[amount]　　

C++：

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int coin : coins) {
            for (int i = coin; i <= amount; ++i) {
                dp[i] += dp[i - coin];
            }
        }
        return dp[amount];
    }
};

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