尝试在 Go 中实现 Java Guava sets.difference

Question

Java番石榴Sets.difference行为：Known = ["v1","v2"]; Incoming = ["v2","v3","v4"]incoming = ["v2","v3","v4"]; knownUpdated = ["v2"]Sets.difference(Known, Incoming) = v1 (To be removed)Sets.difference(incoming, knownUpdated) = v3,v4 (To be added)我在 Go 中尝试过的有以下区别：Output := [v1 v3 v4] (known, Incoming) func Difference(slice1 []string, slice2 []string) []string {    var diff []string    for i := 0; i < 2; i++ {        for _, s1 := range slice1 {            found := false            for _, s2 := range slice2 {                if s1 == s2 {                    found = true                    break                }            }            if !found {                diff = append(diff, s1)            }        }        if i == 0 {            slice1, slice2 = slice2, slice1        }    }    return diff}它给出了对称的差异，但我需要 Guava sets.difference 的行为。我知道我的函数有问题。来自番石榴文档public static Sets.SetView difference(Set set1, Set set2)：返回的集合包含由 set1 包含但由 set2 不包含的所有元素

Answer 1

最直接和易于理解的解决方案是使用映射 - 如果您只使用键，丢弃值，它们与许多其他集合实现（O(1)查找*，唯一键，无序）共享相似的属性。在这一点上，它实际上非常简单：

func Difference(slice1 []string, slice2 []string) []string {

    // Create proper "set" (Maps have unordered pairs and the keys are unique;

    // lookup to check whether value is present is O(1), similar to other

    // implementations)

    m := make(map[string]struct{}, len(slice1))

    for _, el := range slice1 {

        m[el] = struct{}{}

    }

    // Remove values from slice1 present in slice2

    for _, el := range slice2 {

        delete(m, el)

    }

    // Note that res will be non-deterministic: the order of iteration over maps

    // is made random on purpose by Go itself

    res := make([]string, 0, len(m))

    for k := range m {

        res = append(res, k)

    }

    return res

}