Dynamic Programming - making change

问题:

I'm having trouble figuring out my last section of code for a Dynamic Coin Changing Problem. I have included the code below.

I can't figure out the last else. Should I just use the greedy algorithm at that point or can I calculate the answer from values already in the table? I've worked hard on trying to understand this problem and I think I'm pretty close. The method finds the minimum amount of coins needed to make a certain amout of change by creating a table and using the results that are stored in the table to solve the larger problem without using recursion.

public static int minCoins(int[] denom, int targetAmount){
    int denomPosition; // Position in denom[] where the first spot
                       // is the largest coin and includes every coin
                       // smaller.
    int currentAmount; // The Amount of money that needs to be made
                       // remainingAmount <= initalAmount
    int[][] table = new int[denom.length][targetAmount+1];
    for(denomPosition = denom.length-1 ; denomPosition >= 0 ; denomPosition--) {
        for(currentAmount = 0 ; currentAmount <= targetAmount ; currentAmount++){
            if (denomPosition == denom.length-1){
                table[denomPosition][currentAmount] = 
                     currentAmount/denom[denomPosition];
            }
            else if (currentAmount<denom[denomPosition]){
                table[denomPosition][currentAmount] = 
                     table[denomPosition+1][currentAmount];
            }
            else{           
                table[denomPosition][currentAmount] = 
                     table[denomPosition+1][currentAmount]-
                     table[denomPosition][denom[denomPosition]]-1;
            }
        }
    }
    return table[0][targetAmount];
}

回答1:

You don't need to switch to a greedy algorithm for solving the coin changing problem, you can solve it with a dynamic programming algorithm. For instance, like this:

public int minChange(int[] denom, int targetAmount) {

    int actualAmount;
    int m = denom.length+1;
    int n = targetAmount + 1;
    int inf = Integer.MAX_VALUE-1;

    int[][] table = new int[m][n];
    for (int j = 1; j < n; j++)
        table[0][j] = inf;

    for (int denomPosition = 1; denomPosition < m; denomPosition++) {
        for (int currentAmount = 1; currentAmount < n; currentAmount++) {
            if (currentAmount - denom[denomPosition-1] >= 0)
                actualAmount = table[denomPosition][currentAmount - denom[denomPosition-1]];
            else
                actualAmount = inf;
            table[denomPosition][currentAmount] = Math.min(table[denomPosition-1][currentAmount], 1 + actualAmount);
        }
    }

    return table[m-1][n-1];

}

回答2:

//this works perfectly ...

 public int minChange(int[] denom, int targetAmount) 
    {

    int actualAmount;
    int m = denom.length+1;
    int n = targetAmount + 1;
    int inf = Integer.MAX_VALUE-1;

    int[][] table = new int[m][n];
    for (int j = 1; j < n; j++)
        table[0][j] = inf;

    for (int i = 1; i < m; i++) //i denotes denominationIndex
    {
        for (int j = 1; j < n; j++) //j denotes current Amount
        {
            if (j - denom[i-1] >= 0)//take this denomination value and subtract this value from Current amount

                table[i][j] = Math.min(table[i-1][j], 1 + table[i][j - denom[i-1]]);

            else
                table[i][j] = table[i-1][j];

        }
    }




    //display array
        System.out.println("----------------Displaying the 2-D Matrix(denominations and amount)----------------");
        for (int i = 0; i < m; i++) 
        {
            System.out.println("   ");
            for (int j = 0; j < n; j++) 
            {
                System.out.print("  "+table[i][j]);

            }
            System.out.println("   ");
        }

    return table[m-1][n-1];

}

回答3:

Are you over thinking this? If we were trying to give 68 cents change using U.S. coins…

Would ‘denom’ be { 25, 10, 5, 1 } ?

And wouldn’t the answer be “2 quarters, 1 dime, 1 nickel, and 3 pennies” = ‘2 + 1 + 1 + 3 = 7’? So the function should return the value 7. Right?

回答4:

This is actually the correct version of this algorithm.

public static int minChange(int[] denom, int targetAmount) {
    int actualAmount;
    int m = denom.length + 1;
    int n = targetAmount + 1;
    int inf = Integer.MAX_VALUE - 1;

    int[][] table = new int[m][n];
    for(int i = 0; i< m; ++i) {
        for (int j = 1; j < n; j++) {
            table[i][j] = inf;
        }
    }

    for (int denomPosition = 1; denomPosition < m; denomPosition++) {
        for (int currentAmount = 1; currentAmount < n; currentAmount++) {
            if (denom[denomPosition-1] <= currentAmount) {
                // take
                actualAmount = table[denomPosition][currentAmount - denom[denomPosition-1]];
            }
            else {
                actualAmount = inf;
            }                                              // do not take
            table[denomPosition][currentAmount] = Math.min(table[denomPosition-1][currentAmount], 1 + actualAmount);
        }
    }

    return table[m-1][n-1];
}