Consider again the singly linked list(code provided). Write a method called concatenateLists that takes two linked lists A and B as parameters and returns a single list which consists of (in order) the nodes of A followed by the nodes of B. Your method should be able to handle empty and null lists.
Method signature: public static SLinkedList concatenateLists(SLinkedList A, SLinkedList B)
2. Write a method called printRepeats that takes an integer array of size n. This array will consist solely of integers between 1 and n-1 inclusive. It will print out any integer that repeats. Thus, if your input is {5,1,3,5,4,2}, your method will print out 5. Your method must run in O(n) time (and it is up to you to satisfy yourself that it does so).
Method signature: public static void printRepeats(int[] arr)
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Farah Saif
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Hi there
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Duc Nguyen
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Hello,
I am ready to start working on your project.
Q1: Pseudo Code:
public static SLinkedList concatenateLists(SLinkedList A, SLinkedList B) {
1. If first linked list is empty or null return head of second linkedlist
2. else If second linked list is empty or null return head of first linkedlist
3. else store the address of the starting node of the first linked list in a pointer variable , say p
4. Make p the last node of the linked list through simple linked list traversal technique.
5. Store the address of the first node of the second linked list in the next field of the node pointed by p.
6. Return head1.
Q2: Pseudo code
public static void printRepeats(int[] arr){
1. if arr is null or empty print "Array is null"
2. else we will make create two HashSets : duplicates and notDuplicates
3. start loop over array length
4. if notDuplicate HashSet does not contain array entry. Add this array value to not duplicate HashSet
5. else add array value to duplicate HashSet
6. end loop
7. print duplicate HashSet Values
waiting for your response
Thanks
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I am Oracle Certified Java programmer. I can help you to figure out this problem. As I have experience of working in the industry, I have give 2 days deadline.
