Leetcode21_merge_two_sorted_linked_lists.js

var mergeTwoLists = function(l1, l2) {
    // If the first list is null then return the second list
    // Because there is nothing to compare with the second list
    if (!l1) {
        return l2;
    }
    // If the second list is null then return the first list
    // Because there is nothing to compare with the first list
    if (!l2) {
        return l1;
    }
    // This is the head of the new linked list which we will return
    let head = null;
    // We will use this "temp" variable to traverse the new list
    let temp = head;
    // Check for the head of the new list
    // The new head will be chosen from the head of the list which has a smaller value
    if(l1.val < l2.val) {
        // Since there is only one node until now, both head and temp will be equal
        temp = head = new ListNode(l1.val);
        l1 = l1.next;
    } else {
        temp = head = new ListNode(l2.val);
        l2 = l2.next;
    }
    // Loop until one of the list has elements.
    // Even if elements in one of the list are finished, we will come out of the loop
    while (l1 && l2) {
        // Now, create a new node and store the data from list 1 or list 2
        // depending on which list's node has the smaller value
        if (l1.val < l2.val) {
            // New node is created which will be the next of the current temp
            // and it will store the data from the list 1's node
            temp.next = new ListNode(l1.val);
            // Since we have taken node from list 1, we will move to its next node
            l1 = l1.next;
            // Move temp to the current position so that we can add nodes further
            temp = temp.next;
        } else {
            // New node is created which will be the next of the current temp
            // and it will store the data from the list 2's node
            temp.next = new ListNode(l2.val);
            // Since we have taken node from list 2, we will move to its next node
            l2 = l2.next;
            // Move temp to the current position so that we can add nodes further
            temp = temp.next;
        }
    }
    // This loop will check if we have nodes left in the list 1
    // And add all the remaining nodes in the new list
    while (l1) {
        temp.next = new ListNode(l1.val);
        l1 = l1.next;
        temp = temp.next;
    }
    // This loop will check if we have nodes left in the list 2
    // And add all the remaining nodes in the new list
    while (l2) {
        temp.next = new ListNode(l2.val);
        l2 = l2.next;
        temp = temp.next;
    }
    // At the end, we will return the head of the new list
    return head;
};