找到所有快乐的数字，直到

Question

My assignment is to print all the happy numbers from 10 until the program prints 3 consecutive numbers. Those 3 numbers should be 1880 1881 and 1882. i wrote while(num2<1883) just for the loop to stop. i should compare the 3 last "sums" and stop the loop when they are consecutive.

do {
        num = num2;
        while (num > 0 || sum > 9) {
            if (num == 0) {
                num = sum;
                sum = 0;
            }
            sum += Math.pow(num % 10, 2);
            num /= 10;
        }
        if (sum == 1) {
            counter++;
            System.out.println(counter + ") " + num2 + " is a happy number :-)");
        }
        num2++;
        sum = 0;
    } while (num2 < 1883);//<---?????

Answer 1

You should have a List that contains consecutive found numbers. And loop until the length of this list is 3. Inside the loop check if your candidate is an happy number. If it is add it to the list. If it isn't clear the list since you haven't reached a consecutive happy number. After exiting the loop the List contains the three consecutive happy numbers found.

List<Integer> foundNumbers = new ArrayList<Integer>();
int candidate = 10;
while(foundNumbers.length < 3) {
    if(isHappyNumber(candidate)) {
        foundNumbers.add(candidate);
        System.out.println(candidate + " is a happy number");
    } else {
        foundNumbers.clear();
    }
    candidate ++;
}
//Print the list members

Answer 2

The general format is:

int lastHappyNumber = 0;
int currentHappyNumber = 0;
int thirdLastSum = 0;
int secondLastSum = 0;
int firstLastSum = 0;

do {
    currentHappyNumber = getNextHappyNumber();     // <--- fill in with your logic
    thirdLastSum = secondLastSum;
    secondLastSum = firstLastSum;
    firstLastSum = lastHappyNumber + currentHappyNumber;
    lastHappyNumber = currentHappyNumber;
} 
while ((thirdLastSum + 1) == secondLastSum && (secondLastSum + 1) == firstLastSum);

Since this is an assignment, I will leave computing the happy numbers to you, but the idea is that every iteration, you shift down the last 3 sums and then compare them. Once you find a valid solution (where the last three sums were consecutive integers), the value of lastHappyNumber holds the happy number that solves the problem. lastHappyNumber = currentHappyNumber ensures that during the next iteration, the last happy number will be stored and allow for a sum computation to be performed.