Golang：将“切片”作为参考问题传递

Question

I'm trying to write a program that counts inversions within an array, but my array is not being sorted properly due to reference issues and thus messes up my count even though I thought slices were passed by reference in Golang.

Here is my code:

package main

import (
    "fmt"
)

func InversionCount(a []int) int {
    if len(a) <= 1 {
        return 0
    }
    mid := len(a) / 2
    left := a[:mid]
    right := a[mid:]
    leftCount := InversionCount(left) //not being sorted properly due to reference issues 
    rightCount := InversionCount(right) //not being sorted properly due to reference issues

    res := make([]int, 0, len(right)+len(left)) //temp slice to hold the sorted left side and right side

    iCount := mergeCount(left, right, &res)

    a = res        //assigns the original slice with the temp slice values
    fmt.Println(a) //a in the end is not sorted properly for most cases 
    return iCount + leftCount + rightCount
}

    func mergeCount(left, right []int, res *[]int) int {
        count := 0

        for len(left) > 0 || len(right) > 0 {
            if len(left) == 0 {
                *res = append(*res, right...)
                break
            }
            if len(right) == 0 {
                *res = append(*res, left...)
                break
            }
        if left[0] <= right[0] {
            *res = append(*res, left[0])
            left = left[1:]
        } else { //Inversion has been found
            count += len(left)
            *res = append(*res, right[0])
            right = right[1:]
        }
    }

    return count
}

func main() {
    test := []int{4,2,3,1,5}
    fmt.Print(InversionCount(test))
}

What would be the best possible way to solve this problem? I have tried to do something similar to what I did to the res array by forcing the mergeCountfunction to take in a reference of the array, but it seems very messy and it will give me errors.

Answer 1

You either have to pass a pointer to your slice like:

func InversionCount(a *[]int) int {
    if len(*a) <= 1 {
        return 0
    }
    mid := len(*a) / 2
    left := (*a)[:mid]
    right := (*a)[mid:]
    leftCount := InversionCount(&left)   //not being sorted properly due to reference issues
    rightCount := InversionCount(&right) //not being sorted properly due to reference issues

    res := make([]int, 0, len(right)+len(left)) //temp slice to hold the sorted left side and right side

    iCount := mergeCount(left, right, &res)

    *a = res
    fmt.Println(a) //a in the end is not sorted properly for most cases
    return iCount + leftCount + rightCount
}

<kbd>playground</kbd>

Or use copy and change a = res to copy(a, res).

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