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

題解：

第一次WA：

想法是遍歷coins，然後降序dp，每個dp[j]都在dp[j - coins[i]]上多增加了一枚硬幣。

但是沒有考慮到是否可取到問題，即dp[j - coins[i]]是否存在。

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

第一次一直沒想通dp[0] = 1是為什麼：dp[0]理解為組成0的各個硬幣種類，只有一種情況：一枚硬幣不拿，那麼dp[0] = 1。

之後狀態轉移方程為：遍歷coins，dp[j] += dp[j - coins[i]]。

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