LeetCode #518 - Coin Change 2

最新推荐文章于 2020-01-28 00:39:38 发布

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本文探讨了如何计算使用不同面额的无限数量硬币来组成特定金额的组合数。通过示例说明了在给定金额和硬币类型的情况下，如何找出所有可能的组合，并提供了一个基于动态规划的解决方案。

题目描述：

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

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        //dp[i][j]表示利用前i种硬币得到j元钱的组合总数
        vector<vector<int>> dp(coins.size()+1,vector<int>(amount+1,0));
        dp[0][0]=1;
        for(int i=1;i<=coins.size();i++)
        {
            dp[i][0]=1;
            for(int j=1;j<=amount;j++)
            {
                //新加入一种硬币，也可以不用
                dp[i][j]+=dp[i-1][j];
                //当j>=coins[i-1]，dp[i][j]可以由dp[i][j-coins[i-1]]递推
                //dp[i][j-coins[i-1]]可以由dp[i][j-2*coins[i-1]]递推
                if(j>=coins[i-1]) dp[i][j]+=dp[i][j-coins[i-1]];
            }
        }
        return dp[coins.size()][amount];
    }
};