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

To find the standard deviation of all nodes in a linked list, we first need to calculate the mean of all nodes in the list. We'll then iterate through the list again, calculating the sum of the squared differences between each node and the mean. Finally, we'll divide that sum by the total number of nodes in the list minus one, and take the square root of the result.

Here's a Go function that implements this algorithm for a singly linked list:

main.go
package main

import (
    "fmt"
    "math"
)

type Node struct {
    Val  float64
    Next *Node
}

func StdDev(head *Node) float64 {
    if head == nil {
        return 0
    }
    sum, count := 0.0, 0
    curr := head
    for curr != nil {
        sum += curr.Val
        count++
        curr = curr.Next
    }
    mean := sum / float64(count)
    variancesum := 0.0
    curr = head
    for curr != nil {
        squaredDiff := math.Pow(curr.Val-mean, 2)
        variancesum += squaredDiff
        curr = curr.Next
    }
    variance := variancesum / float64(count-1)
    stdDev := math.Sqrt(variance)
    return stdDev
}
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This function takes the head of a linked list as input and returns the standard deviation of all nodes in the list as a float64. Note that we're using the math package to compute the square root and power operations necessary for the standard deviation calculation.