Winner of the Linked List Game

class Solution {
public:
    string determineWinner(ListNode* start) {
        ListNode* traverseNode = start;
        int scoreDiff = 0;
    
        while (traverseNode != nullptr) {
            if (traverseNode->next != nullptr) {
                scoreDiff += (traverseNode->val > traverseNode->next->val) ? 1 : -1;
            }
            traverseNode = traverseNode->next->next; // Move to the next applicable node
        }
    
        if (scoreDiff < 0)
            return "Odd";
        else if (scoreDiff > 0)
            return "Even";
        else
            return "Tie";
    }
};

The given C++ program defines a class Solution with a function determineWinner which determines the winner of a game played with a linked list where nodes represent game elements. Here's a step-by-step breakdown of how the function works:

1. Initialize a pointer traverseNode to the start of the linked list and an integer scoreDiff to zero to track the difference in scores.
2. Iterate through the linked list using a while loop which continues until traverseNode becomes nullptr.
3. Inside the loop, if the next node (traverseNode->next) is not nullptr, compare the current node's value (traverseNode->val) with the next node’s value:
   • If the current node’s value is greater, increase scoreDiff by 1.
   • Otherwise, decrease scoreDiff by 1.
4. Move traverseNode two nodes forward each time (skipping one node) to only check every second node, simulating turn-based scoring.
5. After the loop, determine the winner based on scoreDiff:
   • If scoreDiff is less than 0, return "Odd".
   • If scoreDiff is greater than 0, return "Even".
   • If scoreDiff is 0, return "Tie".

By the end of the function, receive a string indicating the winner ("Odd", "Even", or "Tie") based on the scoring differences computed between adjacent nodes in the linked list. This provides a simple mechanism to evaluate winner scenarios in scenarios where values in a sequence (the linked list) are competed against their immediate neighbors.

class Solution {
    public String calculateWinner(ListNode root) {
        ListNode current = root;
        int scoreDiff = 0;
    
        while (current != null) {
            if (current.next != null) {
                scoreDiff += (current.val > current.next.val) ? 1 : -1;
            }
            current = current.next.next;
        }
    
        if (scoreDiff < 0)
            return "Odd";
        else if (scoreDiff > 0)
            return "Even";
        else
            return "Tie";
    }
}

The Java solution provided is designed to determine the winner of a game played using a linked list where nodes represent game scores. The logic to determine the "Winner of the Linked List Game" involves iterating through the linked list to compare the values of adjacent nodes. Below illustrates how the solution functions:

• Initialize the traversal of the linked list starting from the root.
• Use a variable scoreDiff to keep track of the score difference between consecutive nodes.
• Loop through the linked list:
  • For each node, if it has a next node, update the scoreDiff based on whether the current node's value is greater than the next node's value. Increment scoreDiff by 1 if the current is greater, otherwise decrement by 1.
  • Skip to the node after the next one (current.next.next) in each iteration.
• After completing the traversal, determine the winner:
  • If scoreDiff is negative, return "Odd".
  • If scoreDiff is positive, return "Even".
  • If scoreDiff is zero, return "Tie".

This solution ensures that each pair of consecutive nodes are compared once and that the overall winner is based on the cumulative score differences, effectively deducing if the "even"-positioned or "odd"-positioned nodes generally hold higher values, or if it’s a tie. The approach used capitalizes on linked list traversal and conditional checks to solve the problem efficiently.

class Solution:
    def calculateWinner(self, start_node: Optional[ListNode]) -> str:
        node_pointer, score_margin = start_node, 0
    
        while node_pointer:
            score_margin += (node_pointer.val > node_pointer.next.val) - (node_pointer.val < node_pointer.next.val)
            node_pointer = node_pointer.next.next
    
        if score_margin > 0:
            return "Even"
        elif score_margin < 0:
            return "Odd"
        else:
            return "Tie"

The provided Python function calculateWinner determines the outcome of a linked list game where two players labeled 'Even' and 'Odd' alternate turns to compare the values of linked list nodes. The methodology applied in this function is as follows:

• The function accepts a starting node of a linked list.
• Traverses through the list with the help of a pointer, analyzing the node values in pairs - one node per each player per turn.
• Calculates a score margin by incrementing for 'Even' if the value of the current node (node_pointer.val) is greater than the next node (node_pointer.next.val), and decrementing for 'Odd' if less.
• Skips every second node to simulate the turns of the two players by moving the pointer two nodes forward each iteration.
• After the traversal, evaluates the accumulated score_margin:
  • Returns "Even" if the score margin is positive, indicating player Even scored higher.
  • Returns "Odd" if it is negative, meaning player Odd scored higher.
  • Returns "Tie" if the score is zero, indicating an equal score between players.

Ensure the linked list has an even number of nodes to avoid errors during the game, as the logic assumes pairs of nodes are present for comparison between players. If input linked lists are not formatted as expected, additional checks might be required.