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, int[] coins) { int[] dp = new int[amount + 1]; dp[0] = 1; for (int j = 0; j < coins.length; j++) { for (int i = 1; i <= amount; i++) { if (i - coins[j] >= 0) { dp[i] += dp[i - coins[j]]; } } } return dp[amount]; } }
dp[i] = dp[i - coins[0]] + dp[i - coins[1]] + ... + dp[i - coins[coins.length - 1]] if i - coins[0] >= 0
意思是从dp[i - coins[0]] , dp[i - coins[1]] 是走到前一步的dp加上当前的coin一气呵成