518. Coin Change 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.

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

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给定一堆不同面值的硬币，问凑齐amount块钱，有多少种不同的凑齐方案。

与LeetCode Coin Change类似，也用动态规划。假设dp[i][j]表示用前i类硬币，凑齐j块钱的方案数。则有两种情况，一是不用第i类硬币，有dp[i-1][j]种情况；二是用第i类硬币，则还需要凑齐j-coins[i]块钱，因为每类硬币无穷个，所以还是用前i类硬币凑齐j-coins[i]块钱，即有dp[i][j-coins[i]]种情况。两种情况数相加即可。完整代码如下：LeetCode Coin Change与

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

本代码提交AC，用时60MS。