这段简单的代码的复杂性是多少？

I'm pasting this text from an ebook I have. It says the complexity if O(n²) and also gives an explanation for it, but I fail to see how.

问题：这段代码的运行时间是多少？

public String makeSentence(String[] words) {
    StringBuffer sentence = new StringBuffer();
    for (String w : words) sentence.append(w);
    return sentence.toString();
}

书中给出的答案是：

O(n²), where n is the number of letters in sentence. Here’s why: each time you append a string to sentence, you create a copy of sentence and run through all the letters in sentence to copy them over If you have to iterate through up to n characters each time in the loop, and you’re looping at least n times, that gives you an O(n²) run time. Ouch!

有人可以更清楚地解释这个答案吗？

这似乎是一个误导的问题，因为我刚才正好读到那本书。书中这部分文字是错字！这是上下文：

=================================================== =================

问题：这段代码的运行时间是多少？

1 public String makeSentence(String[] words) {
2 StringBuffer sentence = new StringBuffer();
3 for (String w : words) sentence.append(w);
4 return sentence.toString();
5 }

Answer: O(n²), where n is the number of letters in sentence. Here’s why: each time you append a string to sentence, you create a copy of sentence and run through all the letters in sentence to copy them over. If you have to iterate through up to n characters each time in the loop, and you’re looping at least n times, that gives you an O(n²) run time. Ouch! With StringBuffer (or StringBuilder) can help you avoid this problem.

1 public String makeSentence(String[] words) {
2 StringBuffer sentence = new StringBuffer();
3 for (String w : words) sentence.append(w);
4 return sentence.toString();
5 }

=================================================== ===================

Have you noticed that the author messed it up? The O(n²) solution she mentioned (the first one) was exactly the same as the 'optimized' one (the latter). So, my conclusion is that the author was trying to render something else, such as always copying the old sentence to a new buffer when appending every next string, as the example of an O(n²) algorithm. StringBuffer should not be so silly, as the author also mentioned 'With StringBuffer (or StringBuilder) can help you avoid this problem'.