Change exchange II - [leetcode]

class Solution {
    
    // give an example amount = 5, coins = [1, 2, 5]
    public int change(int amount, int[] coins) {
    
        // Recursive expression 
        int[] dp = new int[amount + 1];
        // initialization dp Array , Indicates that the amount is 0 There is only one case , That is to pretend nothing 
        dp[0] = 1;
        for(int i = 0;i < coins.length; i++) {
    // First traversal coins

            //dp[j]： Make up the total amount j The number of currency combinations is dp[j]
            // First round outer layer for loop , Show consideration coins[0]==1
            // Make up the total amount 1 The number of currency combinations is dp[1] = dp[1] + dp[0]; ( Is to consider 1 And don't consider 1 The sum of ) At this point, you may think dp[1] What is the value , Actually dp When the array is initialized , It's worth it all 0 了 , So here dp[1] = dp[1] + dp[0] = 0 + 1 = 1
            // When j++, after dp[2] = dp[2] + dp[2-1];（ Is to consider 2 And don't consider 2 The sum of ）
            // Why initialize here j=coins[i]??
            // consider j=coins[1] The situation of ： here j==2, be dp[2] = dp[2] + dp[2-2];dp[3] = dp[3] + dp[3-2];
            // Be careful ： You may think of dp[3-2] What is the value ？ yes 0 Do you ？ No , Actually dp The array is in the last round for It's already right when you cycle dp The array is given a new value , Second round outer layer for The loop just changes the value based on the previous round ,dp An array is a dynamically changing array 
            for(int j = coins[i]; j <= amount; j++) {
    
                dp[j] += dp[j - coins[i]];
            }

// // Print dp Array 
// for(int i1 = 0; i1 < dp.length;i1++) {
    
// System.out.print(dp[i1]);
// }
// System.out.println();
        }

        return dp[amount];
    }
}