find the standard deviation of all nodes in a linked list in typescript

To find the standard deviation of all nodes in a linked list, we can first calculate the mean of all the node values. Then, we can iterate through the list again to calculate the sum of the squared differences between each node value and the mean. Finally, we can divide the sum of squared differences by the number of nodes and take the square root to get the standard deviation.

Here's an implementation of a function in TypeScript that takes a linked list as input and returns the standard deviation of all the nodes:

index.ts
function findLinkedListStandardDeviation(head: ListNode): number {
  let current: ListNode | null = head;

  // Calculate the mean
  let sum = 0;
  let count = 0;
  while (current != null) {
    sum += current.val;
    count++;
    current = current.next;
  }
  const mean = sum / count;

  // Calculate the sum of squared differences
  let squaredDiffsSum = 0;
  current = head;
  while (current != null) {
    squaredDiffsSum += (current.val - mean) ** 2;
    current = current.next;
  }

  // Calculate the standard deviation
  const variance = squaredDiffsSum / count;
  const standardDeviation = Math.sqrt(variance);
  return standardDeviation;
}
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Note that this assumes that the linked list nodes have a val property that holds the node value. Also, the function assumes that the list has at least one node; you may need to add additional input validation if this is not always the case.