Answer:
- import java.util.ArrayList;
- public class Main {
-
- public static void main(String[] args) {
- ArrayList<String> strList =new ArrayList<String>();
- strList.add("to");
- strList.add("be");
- strList.add("or");
- strList.add("not");
- strList.add("to");
- strList.add("be");
- strList.add("hamlet");
-
- swapPairs(strList);
- System.out.println(strList);
- }
-
- public static void swapPairs(ArrayList<String> list){
- for(int i=0; i < list.size()-1; i+=2){
- String temp = list.get(i);
- list.set(i, list.get(i+1));
- list.set(i+1, temp);
- }
- }
- }
Explanation:
Firstly, let's create a method swapPairs that take one ArrayList (Line 18). In the method, use a for-loop to traverse through each item in the ArrayList and swap the items between the current items at index-i and at index-i+1 (Line 19-22). The index-i is incremented by two in next loop and therefore the next swapping will proceed with third and fourth items and so forth.
In the main program, create a sample ArrayList (Line 5-12) and then test the method (Line 14) and print the output (Line 15). We shall get [be, to, not, or, be, to, hamlet].
Answer: I can't really code a whole thing for you but use VS Code for this it'll make your life easier in the long run.
Perhaps instead of cubicles, desks are organized in an open workspace which promotes collaboration (and makes it easier).
Answer:
B.O(n).
Explanation:
Since the time complexity of visiting a node is O(1) in iterative implementation.So the time complexity of visiting every single node in binary tree is O(n).We can use level order traversal of a binary tree using a queue.Which can visit every node in O(n) time.Level order traversal do it in a single loop without doing any extra traversal.
