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

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

标签：int,coins,number,518,amount,Change,Input,Coin,dp
来源： https://www.cnblogs.com/wentiliangkaihua/p/12231857.html