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CS1020 Lab 3 (Linked List). Problem 1 : Balls Problem 2 : Josephine Problem 3 : Alternating List . Problem 1: Balls. #1: Balls. Problem: Given N balls labeled from 1 to N and M operations, determine the last state after doing M operations. Input The first line contains two integers,
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CS1020 Lab 3(Linked List) Problem 1 : Balls Problem 2 : Josephine Problem 3 : Alternating List
#1: Balls • Problem: • Given N balls labeled from 1 to N and M operations, determine the last state after doing M operations. • Input • The first line contains two integers, • N (1<= N <= 1,000) and M (1<= M <= 1,000) • The next M lines contain the operations. Sample Input 10 5 A 2 1 A 10 1 A 5 6 B 6 9 R 3
Operations: • A x y: move the ball labeled with number x to the left of the ball labeled with number y. • B x y: move the ball labeled with number x to the right of the ball labeled with number y. • R x: Remove the ball labeled with number x from our list. Output: • Output the final arrangement of the N balls from left to right. Each number is followed by a whitespace.
Discussion (1) 1. What kind of data structure can we use to solve this problem?
Discussion (2) • Hint: use List • Q: What kind of List (Singly-Linked List/ Doubly-Linked List/ Circular-Linked List)? • A: Use Doubly-Linked List Why?
Discussion (3) 2. Simplest implementation: -Traversing all the nodes (to find the correct node) for removing and moving Better implementation: - Without traversing through the nodes(will be discussed later)
Discussion (4) ListNode stores: 1. Number/Id 2. Previous Ball 3. Next Ball int ID ID ListNodeprev ListNode next
Discussion (5) Remove: 1. Find position of x 2. Update the affected ListNode.
Removing • Rough steps: • Iterate though the list to find the node with label x, and remember which node it is (point to it from a variable) • Adjust the pointers
Adjusting pointers for removing • Remove node X from position Previous of X X Next of X prev prev prev next next next
Adjusting pointers for removing • Remove node X from position Previous of X X Next of X prev prev prev next next next X.next.prev = X.prev X.prev.next = X.next Must handle special case of 1stand last nodeswhen X.prev == null or X.next == null
Moving • Rough steps: • Iterate though the list to find the node with labels x and y, and store them in two variables. • Remove node X • Insert node X left/right of node Y • (practice using back the same node removed and insert it back, rather than creating a new node)
Insert node • Insert X on the left of Y Previous of Y Y prev prev next next
Insert node • Insert X on the left of Y Previous of Y X Y prev prev prev next next next Do it in the right order – don’t “lose” the previous of Y node X.prev = Y.prev X.next = Y Y.prev.next = X Y.prev = X Must handle special case of inserting X before the 1st node when Y.prev == null
Insert node • Insert X on the left of Y Previous of Y X Y prev prev prev next next next Do it in the right order – don’t “lose” the previous of Y node X.prev = Y.prev X.next = Y Y.prev.next = X Y.prev = X Must handle special case of inserting X before the 1st node when Y.prev == null
Insert node • Insert X to the right of Y Y Next of Y prev prev next next
Insert node • Insert X to the right of Y Y X Next of Y prev prev prev next next next Do it in the right order – don’t “lose” the previous of Y node X.next = Y.next X.prev = Y Y.next.prev = X Y.next = X Must handle special case of inserting X after the last node when Y.next == null
Insert node • Insert X to the right of Y Y X Next of Y prev prev prev next next next Do it in the right order – don’t “lose” the previous of Y node X.next = Y.next X.prev = Y Y.next.prev = X Y.next = X Must handle special case of inserting X after the last node when Y.next == null
Discussion (7) An idea to improve performance: • We store a “reference” of all balls (ListNodes) in an array, so that we don’t need to traverse the Linked List to find the balls every time. head 1 2 4 3 5
#2: Josephine • Problem: • Given N candidates in circle, we want to keep removing the K-th candidate until we find the number of people in the circle equals to 1. Output the removed candidate for each remove operation.
#2: Josephine • Input • The first line consists of T, the number of test cases, T <= 100. • The following T lines describe T test cases, each line containing two integers, N and K. • Output • Output the final arrangement of the N balls from left to right. Each number is followed by a whitespace. Sample Input 2 3 1 4 2 Sample Output 1 2 3 2 4 3 1
Discussion (1) 1. What kind of data structure can we use to solve this problem?
Discussion (2) • Hint:- Use Circular Linked List Why? Simulates the problem!
Discussion (3) • Keep removing K-th person by iterating K times from current position. • Update the State of ListNode. • More or less similar to Previous Problem but using different type of LinkedList.
Circular linked list • Removing a node: • Since you don’t have the previous pointer in a singly-linked circular linked list, you need to keep track of the previous node when you traverse the list! head 1 2 4 3 5
#3: Alternating list Problem: • Given a list of integers, and a list of operations, determine whether it is alternating after each operation • Definition of Alternating: -Adjacent integers have different signs • List of one or zero elements is alternating
#3: Alternating list Input • 1 <= N <= 100 1 <= Q <= 100 • N is the size of the original linked list • Q the number of operations • Output • For each operation print “YES” if the updated linked list is alternating, otherwise print “NO”. Sample Input 4 4 1 -2 3 -4 M 1 3 A 1 1 14 R 2 2 A 2 1 -11 Sample Output YES NO NO YES
Discussion (1) 1. What kind of data structure can we use to solve this problem?
Discussion (2) • Hint: use List • Q: What kind of List (Singly-Linked List/ Doubly-Linked List/ Circular-Linked List)? • A: Use (Singly) Linked List Why?
Discussion (2) - Operations • M [index] [size]Move “size” number of elements starting from index to the end of list M 2 3
Discussion (2) - Operations • Pseudo-code:function Move (int index, int size) { for i=1 to size { temp remove element at indexinsert temp to the end of list}} • Why removing at the index only works?
Discussion (2) - Operations • R [index] [size] Remove “size” number of elements starting from index. R 2 3
Discussion (2) - Operations • Pseudo-code:fun Remove (int index, int size) { for i=1 to size {remove element at index }} • Can you use this to implement Move?
Discussion (2) - Operations • A [index] [size] [value] add the elements between index [index] and [index + size - 1] (inclusive) with [value]. +5 +5 +5 A 2 3 5
Discussion (2) - Operations • Pseudo-code:fun AddValue (int index, int size, int value) { for i=1 to size { temp remove value at index+i-1 insert (temp+value) to index+i-1 }} • How do we get the (index+i-1)? • (or we can cheat and just modify the number in listNode…)
Discussion (3) - Check Alternating • List of 1 element : YES • Loop through list: once two adjacent elements with same sign found, return “NO” i.e. • Keep track of “previousSign”, and then check every iteration if “previousSign == currentSign” • Return “YES” if loop completes without returning
The End Any Questions?