递归找零：如何修改我的算法来打印所有的组合？ [英] Making change recursively: How do I modify my algorithm to print all combinations?

问题描述

我有一种算法，递归使得变化以下述方式：

I have an algorithm that recursively makes change in the following manner:

public static int makeChange(int amount, int currentCoin) {

        //if amount = zero, we are at the bottom of a successful recursion
        if (amount == 0){

            //return 1 to add this successful solution
            return 1;           

            //check to see if we went too far
        }else if(amount < 0){

            //don't count this try if we went too far
            return 0;

            //if we have exhausted our list of coin values
        }else if(currentCoin < 0){              

            return 0;

        }else{  

            int firstWay = makeChange(amount, currentCoin-1);
            int secondWay = makeChange(amount - availableCoins[currentCoin], currentCoin);

            return firstWay + secondWay;            
        }       
}   

不过，我想增加存储或打印的各种组合，因为他们成功返回的能力。我有一个有点很难包装我的头，围绕如何做到这一点。原来的算法是pretty的容易，但现在我很沮丧。有什么建议？

However, I'd like to add the capability to store or print each combination as they successfully return. I'm having a bit of a hard time wrapping my head around how to do this. The original algorithm was pretty easy, but now I am frustrated. Any suggestions?

CB

推荐答案

我用下面的：

/**
     * This is a recursive method that calculates and displays the combinations of the coins included in
     * coinAmounts that sum to amountToBeChanged.
     * 
     * @param coinsUsed is a list of each coin used so far in the total.  If this branch is successful, we will add another coin on it.
     * @param largestCoinUsed is used in the recursion to indicate at which coin we should start trying to add additional ones.
     * @param amountSoFar is used in the recursion to indicate what sum we are currently at.
     * @param amountToChange is the original amount that we are making change for.
     * @return the number of successful attempts that this branch has calculated.
     */private static int change(List<Integer> coinsUsed, Integer currentCoin, Integer amountSoFar, Integer amountToChange)
    {
        //if last added coin took us to the correct sum, we have a winner!
        if (amountSoFar == amountToChange)
        {
            //output 
            System.out.print("Change for "+amountToChange+" = ");

            //run through the list of coins that we have and display each.
            for(Integer count: coinsUsed){

                System.out.print(count + " ");
            }
            System.out.println();       

            //pass this back to be tallied
            return 1;
        }

        /*
         * Check to see if we overshot the amountToBeChanged
         */
        if (amountSoFar > amountToChange)
        {
            //this branch was unsuccessful
            return  0;
        }

        //this holds the sum of the branches that we send below it
        int successes=0;

        // Pass through each coin to be used
        for (Integer coin:coinAmounts)
        {
            //we only want to work on currentCoin and the coins after it
            if (coin >= currentCoin)
            {
                //copy the list so we can branch from it
                List<Integer> copyOfCoinsUsed = new ArrayList<Integer>(coinsUsed);

                //add on one of our current coins
                copyOfCoinsUsed.add(coin);

                //branch and then collect successful attempts
                successes += change(copyOfCoinsUsed, coin, amountSoFar + coin, amountToChange);
            }
        }

        //pass back the current
        return successes;

    } 

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