时间复杂度O（N ^ 2）在这里如何？

Question

I already know that the answer to this question is O(N^2) but I am not able to see how. I know the for loop runs N times, but how can it run N^2 times?

public static String rev(String s) {
    String r = "";
    int N = s.length();
    for (int i = 0; i < N; i++) {
        r = s.charAt(i) + r;
    }
    return r;
}

Answer 1

在Java中， String串联r = s.charAt(i) + r在循环中为O(N^2) ，因为Strings是不可变的-在每个串联中都会创建String的新副本。