Comment From: ZhangShenao
HashSet.toArray() will use the actual size of HashSet, so it's unnecessary to pass a large size array. @jasonab
Comment From: harawata
Thank you, @ZhangShenao !
Comment From: ZhangShenao
HashSet.toArray() will use the actual size of HashSet, so it's unnecessary to pass a large size array. @jasonab
Comment From: harawata
Thank you, @ZhangShenao !