Time Complexity of Stringbuilder Reverse Method

I want to find the time complexity of StringBuilder reverse method. Here the source code of the reverse method:

 public AbstractStringBuilder reverse() {
        boolean hasSurrogate = false;
        int n = count - 1;
        for (int j = (n-1) >> 1; j >= 0; --j) {
            char temp = value[j];
            char temp2 = value[n - j];
            if (!hasSurrogate) {
                hasSurrogate = (temp >= Character.MIN_SURROGATE && temp <= Character.MAX_SURROGATE)
                    || (temp2 >= Character.MIN_SURROGATE && temp2 <= Character.MAX_SURROGATE);
            }
            value[j] = temp2;
            value[n - j] = temp;
        }
        if (hasSurrogate) {
            // Reverse back all valid surrogate pairs
            for (int i = 0; i < count - 1; i++) {
                char c2 = value[i];
                if (Character.isLowSurrogate(c2)) {
                    char c1 = value[i + 1];
                    if (Character.isHighSurrogate(c1)) {
                        value[i++] = c1;
                        value[i] = c2;
                    }
                }
            }
        }
        return this;
    }

Here's a link to the doc: documentation Which is the time complexity?

Is there any way to perform more efficiently the reversion of a String?

2

1 Answer

It is impossible to revert a String in less than O(n) and the algorithm you have posted clearly is O(n) since it contains two successive O(n) loops.

3

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Sarah Jenkins

Sarah Jenkins

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Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.

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