Question

我试图找到总和为给定数字n的最小平方整数。

我用递归函数解决了它，但我想迭代地完成它。
如何使用一些循环而不是递归方法？

 public static ArrayList<Integer> minLen(int n)
    {
        // base case of recursion
        if (n == 0)
            return new ArrayList<Integer>();

        ArrayList<Integer> best = null;
        int bestInt = -1;
        for (int i = 1; i*i <= n; ++i)
        {
            // Check what happens if we use i^2 as part of our representation

            ArrayList<Integer> guess = minLen(n - i*i);
            System.out.println("i:"+i);
            System.out.println("Guess"+guess);
            // If we haven't selected a 'best' yet (best == null)
            // or if our new guess is better than the current choice (guess.size() < best.size())
            // update our choice of best
            if (best == null || guess.size() < best.size())
            {
                best = guess;
                System.out.println("best"+best);

                bestInt = i;
                System.out.println("bestInt"+bestInt);
            }
        }

        best.add(bestInt);
        System.out.println("bestInt"+bestInt);
        System.out.println("best"+best);
        return best;
    }

Answer 1

这可以使用带循环的动态编程解决

您正在解决的问题是找到总和为n的最小平方整数。

可以通过以下伪代码在Dynamic Programming中解决：

int[] sol = new int[n+1];
//init:
sol[0] = 0;
for (int i = 1; i <= n; i++) sol[i] = Integer.MAX_VALUE;
//fill the value for each i in increasing order:
for (int i =1; i <= n; i++)
   for (int j = 1; j*j <= i; j++)
          //make sure sol[i] contains the best possible solution
          sol[i] = Math.min(sol[i], sol[i-j*j] + 1);

上面给出了尽可能少的数字（sol[n]就是答案），以获得平方数字：

List<Integer> numbers = new ArrayList<>();
int curr = n;
//while we haven't 'exhausted' the number:
while (curr > 0) { 
    //check all numbers we could have added
    for (int j=1; j*j <= curr ; j++) {
         if found the correct number we used:
         if (sol[curr  - j*j] == sol[curr ] - 1) { 
               //add it to the solution, reduce it from curr and repeat
               numbers.add(j);
               curr = curr - j*j;
               break;
         }
    }
}

我们的想法是重复您的DP解决方案的步骤，并重复您在此过程中做出的选择。

总和为n的最小平方整数 - 如何迭代地完成？

1 个答案: