检查两个数组是否具有相同成员的最佳方法

Question

I have an array of strings, I need to compare this to another array of strings, but they may be in a different order. What's the best way to compare the two arrays?

This is what I have so far, just wondering if there is a simpler / more efficient way I'm missing.

func unorderedEqual(first, second []string) bool {
    if len(first) != len(second) {
        return false
    }
    for _, value := range first {
        if !contains(value, second) {
            return false
        }
    }
    return true
}

func contains(needle string, haystack []string) bool {
    for _, matchValue := range haystack {
        if matchValue == needle {
            return true
        }
    }
    return false
}

Answer 1

Given that you are doing a length check, I'm going to go with the assumption that implies that they are 1:1, just ordered differently.

You can do this in one pass (each) using a map[string]bool to check existence in both. This utilizes the fact that the map returns the zero value of a bool, which is false, when the key is not present.

Disclaimer: Technically this is order O(n)*O(map). The Go Programming Language Specification does not make any performance guarantees for map types.

https://play.golang.org/p/2LUjN5LkXLL

func unorderedEqual(first, second []string) bool {
    if len(first) != len(second) {
        return false
    }
    exists := make(map[string]bool)
    for _, value := range first {
        exists[value] = true
    }
    for _, value := range second {
        if !exists[value] {
            return false
        }
    }
    return true
}

If you want to get nit-picky about memory usage, you could save yourself storing a bunch of bools (which is usually negligible, but to each their own) by using a map[string]struct{} (the empty struct), and you just check existence a little differently, as in this example.

https://play.golang.org/p/MjwP_wCtYZV

Set

exists[value] = struct{}{}

Check

if _, ok := exists[value]; !ok {
    return false
}