如何将for循环迭代回到它在java中访问的最后一个索引

Question

I have an array of integer, and i am trying to iterate back the for loop to the last index it visited based on some condition , Lets say i have 2 9 15 19 23 28 37 elements in that loop i am giving a condition that if each element of that loop is greater than a number lets say 8 , it will process that element again.

Here my code is

List<Integer> result = new ArrayList<Integer>();
int n = 6;
for (int i = 0; i < n; i++) {
    int h = 8;
    int r = a[i] - h;
    if (r <= 0) {
        result.add(1);
    } else if (r >= 0) {
        result.add(1);
    }
}

Here h is the array which holds the declared elements.r is an integer that checks if elements are greater than the hit element ie 8.So the condition is if the elements are less that h the arraylist will add 1, else the control will go back to the operation int r = a[i] -h, for the same element.For example , 2 is less that 8 the arraylist will add 1 , but for 9 the control will do that same minus operation and come to the else part and add 1 to the arraylist.The last element processed by the loop if not zero will not be added to the list.Is is possible? please help.

Answer 1

You could be returned a step back by --i :

if (a[i] - h > 0) {
    // a[i] is greater than h
    --i; // process a[i] again on the next iteration
}

As @Stefan Warminski noticed it will lead to an infinite loop because we always will process the first element than is greater than the h .

The workaround could be creating an array int[] changes the same length as the origin list and putting a value into an appropriate cell which will indicate how many times changes[i] we process an a[i] element:

if (a[i] - h > 0 && changes[i]++ < N) { ... }

where N is how many times you want to process an element.

Complete code snippet:

int[] a = {2, 9, 15, 19, 23, 28, 37};
int[] changes = new int[a.length];

int h = 8;
int N = 2;

for (int i = 0; i < a.length; i++) {
    if (a[i] - h > 0 && changes[i]++ < N) {
        System.out.println(a[i] + " is processed " + changes[i] + " times");
        --i;
    }
}

Output:

9 is processed 1 times
9 is processed 2 times
15 is processed 1 times
15 is processed 2 times
19 is processed 1 times
19 is processed 2 times
23 is processed 1 times
23 is processed 2 times
28 is processed 1 times
28 is processed 2 times
37 is processed 1 times
37 is processed 2 times

Tip: declare the h variable outside the for statement, it won't change inside (there is no need to create a variable on every iteration).

Answer 2

Note that the test r<=0 is useless : you can just take result.add(1) outside of the tests since you do that operation in any case.

int h=8;
for(int i=0;i<n;i++){
    int r = a[i] -h;
    result.add(1);
    if(r >=0){
        'loop back'
    }
}

Then, if your goal is to "process that element again", do you really need to iterate back ? You already have the element, just process it again.

List<Integer> result = new ArrayList<Integer>();
int n =6;
int h=8;
for(int i=0;i<n;i++){
    int r = a[i] -h;
    result.add(1); // Processed once
    if(r >=0){
        result.add(1); // Processed twice
    }
}

Answer 3

List<Integer> result = new ArrayList<Integer>();
int n =6;
for(int i=0;i<n;i++){
    int h=8;
    int r = a[i] -h;
    if(r <=0){
        result.add(1);
    }else if(r >=0){
        result.add(1); 
        i--;           
    }
}

Using this code you can process the number greater than 8 again.If r>=0 then i is decremented and the numbers greater than 8 is processed again.

Answer 4

If I understand correctly, you want to process an element multiple times, only when that element meets some condition.

You can simply loop over the list and then if the element is greater than h , you just execute the desired action (in my case System.out.println() ) an n number of times.

List<Integer> list = Arrays.asList(2, 9, 15, 19, 23, 28, 37);

int h = 8;
int n = 2;
list.stream().forEach(t -> {
    int r = (t > h ? n : 1);
    for (int i = 0; i < r; i++) {
        System.out.println(t);
    }
});

Or the below-Java-8 version:

List<Integer> list = Arrays.asList(2, 9, 15, 19, 23, 28, 37);

int h = 8;
int n = 2;
for (int t : list) {
    int r = (t > h ? n : 1); // Execute 'n' times if condition is met,
                             // otherwise, execute once
    for (int i = 0; i < r; i++) {
        System.out.println(t); // Code you want to execute
    }
}

---

Besides,

int h = 8;
int r = a[i] - h;
if (r <= 0) {
    result.add(1);
}
else if (r >= 0) {
    result.add(1);
}

doesn't make sense. In fact, you are comparing a[i] with h . The following code has the same effect:

int h = 8;
result.add(1);
if (a[i] == h) {
    // Add the number 1 again to the list
    result.add(1);
}