leetcode-2-add-two-numbers.swift

/**

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https://leetcode.com/problems/two-sum/


Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

*/

/**
Big O Annotation
Time complexity O(n) where n is the combined amount of elements in l1 and l2.
Space complexity O(n) where n is the combined amount of elements in l1 and l2.
*/
func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        
    // Get a mutable reference to each head of l1 and l2
    var currentL1: ListNode? = l1
    var currentL2: ListNode? = l2

    // Create a reference to a node we're going to append on
    var sumList: ListNode? = ListNode(-1)

    // And a reference to its head
    var head: ListNode? = sumList

    // The carry for values > 9
    var carry: Int = 0
    
    // While we haven't fully iterated both lists yet...
    while currentL1 != nil || currentL2 != nil {
        
        /**
        Get a dummy value of 0 for each list
        in the niche case that either list
        is longer than the other.
        */
        var value1: Int = 0
        var value2: Int = 0
        
        // Replace with l1's value if it exists
        if let l1 = currentL1 {
            value1 = l1.val
        }
        
        // Same for l2
        if let l2 = currentL2 {
            value2 = l2.val
        }
        
        // Sum up both values and the last carry
        let sum: Int = value1 + value2 + carry 

        // Get the value < 10
        let value = sum % 10

        // Get the carry if above 10
        carry = sum / 10

        // Add a node with the new value
        sumList?.next = ListNode(value)

        // And move the head reference
        sumList = sumList?.next
        
        // Move l1 and l2 along as well
        currentL1 = currentL1?.next
        currentL2 = currentL2?.next
    }
    
    // If we still have carry left
    if carry > 0 {
        /**
        Add a node for the carry.
        This case could happen on 
        (9 -> 9) + (1) = (0 -> 0 -> 1)
        */
        sumList?.next = ListNode(carry)
    }
    
    // Return head.next to lose the -1 dummy head
    return head?.next
}