StringBuilder反向方法的时间复杂度

Question

我想找到StringBuilder反向方法的时间复杂度.这里是反向方法的源代码:

 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;
    }

这是doc的文档链接:文档 的时间复杂度是多少？

有没有办法更有效地执行String的还原？

Answer 1

在小于O(n)的情况下恢复字符串是不可能的,并且您明确发布的算法是O(n),因为它包含两个连续的O(n)循环.