Java番石榴Sets.difference行為：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 中嘗試過的有以下區(qū)別：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}它給出了對(duì)稱的差異，但我需要 Guava sets.difference 的行為。我知道我的函數(shù)有問題。來自番石榴文檔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
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天涯盡頭無女友
TA貢獻(xiàn)1831條經(jīng)驗(yàn) 獲得超9個(gè)贊
最直接和易于理解的解決方案是使用映射 - 如果您只使用鍵，丟棄值，它們與許多其他集合實(shí)現(xiàn)（O(1)查找*，唯一鍵，無序）共享相似的屬性。在這一點(diǎn)上，它實(shí)際上非常簡(jiǎn)單：
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
}
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