leetcode-203-remove-linked-list-elements.swift

/**

Copyright (c) 2020 David Seek, LLC

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OR OTHER DEALINGS IN THE SOFTWARE.





https://leetcode.com/problems/remove-linked-list-elements/


Remove all elements from a linked list of integers that have value val.

Example:

Input:  1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5

*/

/**
In place approach.
Faster -> Gets us into the 100$ range.
And doesn't use any space.

Big O Annotation
Time complexity O(n) where n is the amount of nodes in head.
Space complexity O(1) where removing in place / not using extra space.
*/
func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
        
    // Return nil if we don't have a head node
    guard let head = head else {
        return nil
    }

    // Get a reference to the head
    var current: ListNode? = head
    
    // Check that the head doesn't need to be removed
    while current != nil && current!.val == val {

        // If so, set the next node as new head
        current = current?.next
    }
    
    // Store a reference to the final head
    var root: ListNode? = current
    
    // While we still have nodes...
    while current != nil {
        
        // Check, that the next one exists and is not .val == val
        while current!.next != nil && current!.next!.val == val {

            /**
            If so, cut the node out and set to next.next
            Will be repeated until we don't have .val == val
            */
            current!.next = current!.next!.next
        }
        
        /**
        We need to make sure, that the final .next.val
        is not .val for niche cases
        */
        if current!.next?.val == val {
            current!.next = nil
        }
        
        // Continue with the next .next
        current = current?.next
    }
    
    return root
}

/**
Stored values approach. 
Takes longer and is less space efficient.

Big O Annotation
Time complexity O(n) where n is the amount of nodes in head.
Space complexity O(n) where n is the amount of nodes in head.
*/
func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
        
    // Get an array of values of the list
    var values: [Int] = []
    head?.traverse { values.append($0) }

    // Filter by the desired removal value
    values = values.filter { $0 != val }

    // Init a new list
    var newHead: ListNode? = ListNode(-1)

    // Store it's head
    var root: ListNode? = newHead
    
    // Iterate through values
    for value in values {

        // And init a new node for every value
        let node = ListNode(value)

        // Set it
        newHead?.next = node

        // And store it as next
        newHead = node
    }

    // Return .next to lose the -1 head
    return root?.next
}

/**
Simple function to traverse 
a list node by node    
*/
extension ListNode {
    func traverse(_ visit: (Int) -> Void) {
        visit(val)
        next?.traverse(visit)
    }
}