Java StringBuilder.substring（）方法的时间复杂度是多少？ 如果它是线性的，是否有一个恒定的时间复杂度方法来获得子串？

Question

In my opinion it should be constant(O(1)) time complexity. However I was told that a new String object has to be instantiated when invoking stringBuilder.substring() method. (It is not static method). If this is true, how can I get a substring within a constant time complexity?

Answer 1

You cannot possibly create a String from a StringBuilder in constant time and have its immutability maintained. Additionally, as of Java 7 Update 25, even String#substring() is linear time because structural sharing actually caused more trouble than it avoided: substring of a huge string retained a reference to the huge char array.

Answer 2

In my opinion it should be constant(O(1)) time complexity.

This is not possible. Since Java strings are immutable, while StringBuilder s are mutable, all operations that produce a String must make a copy.

If this is true, how can I get a substring within a constant time complexity?

You cannot. Without a copy, changing characters inside a substring would mutate the String , which is not allowed.

Answer 3

StringBuilder's substring actually copies the characters from start index to end index, so it tries to array copy of characters of substring length(ie end index-startindex). This will be the time complexity of this operation O(substringlength) as the loop runs from start index to end index which cannot be avoided. With this the substring char[], string object will be instantiated.