Understanding this remove method solution with arrayList [duplicate]

问题

This methods duty is to remove all occurrences of the value toRemove from the arrayList. The remaining elements should just be shifted toward the beginning of the list. (the size will not change.) All "extra" elements at the end (however many occurrences of toRemove were in the list) should just be filled with 0. The method has no return value, and if the list has no elements, it should just have no effect. Cannot use remove() and removeAll() from the ArrayList class.

The method signature is:

public static void removeAll(ArrayList<Integer> list, int toRemove);

The solution:

public static void removeAll(ArrayList<Integer> list, int toRemove) {
    for (int i = 0; i < list.size(); i++) {
        if (list.get(i) = toRemove) {
            for (int j = i + 1; j < list.size(); j++) {
                list.set(j - 1, list.get(j));
            }
            list.set(list.size() - 1, 0);
            i--;
        }
    }

I understand the first for loop and the if statement well. Because one would want to iterate through the entire arrayList one-by-one and for each index with a number present in the arrayList check if it is, in fact the toRemovee integer. After this point I become lost.

Why another for loop? Why are we taking the previous loops variable and adding 1 to it? why within this second loop are we using the parameters "list" and using a set method? why j - 1? why list.get(j)? Why after this second loop is over is there the line: list.set(list. sise () - 1, 0) ? why the i--?

There are a lot of moving parts and I would like to understand the logic.

Thank you

回答1:

Firstly the if statement is assignment operation which isn't correct. You need to change the = to ==. I have explained each step in the code -

public static void removeAll(ArrayList<Integer> list, int toRemove) {
    //start with the first number in the list until the end searching for toRemove's
    for (int i = 0; i < list.size(); i++) {
        //if the value at i is the one we want to remove then we want to shift
        if (list.get(i) == toRemove) {
            //start at the index to the right until the end
            //for every element we want to shift it the element to its left (i == j - 1)
            for (int j = i + 1; j < list.size(); j++) {
                //change every value to whatever was to the right of it
                //this will overwrite all values starting at the index where we found toRemove
                list.set(j - 1, list.get(j));
            }

            //now that everything is shifted to the left, set the last element to a 0
            list.set(list.size() - 1, 0);
            //decrement to adjust for the newly shifted elements
            // this accounts for the case where we have two toRemoves in a row
            i--;
        }
    }
}

By the end of this function any value that matches toRemove is "removed" by shifting the arraylist to the left everytime the value is found and the last element will be set to 0.

Example

removeAll([1,2,3,4,5,5,6,7,8,5,9,5], 5)

Output

[1,2,3,4,6,7,8,5,9,0,0,0]

回答2:

Please see the following to learn details (from minute 5:50 or 5:57)

https://www.youtube.com/watch?v=qTdRJLmnhQM

You need the second for loop ,to take all elements after the element you’ve removed it and shift it over one to the left so basically it’s filling up the space that is left empty from removing it so this is what this is doing.

来源：https://stackoverflow.com/questions/41946923/understanding-this-remove-method-solution-with-arraylist

标签

java

arraylist