charAt如何反转此方法中的输入？

Question

//method
public static String foo(String s)

{
if (s.length() == 1)

return s;

else

return foo(s.substring(1)) + s.charAt(0);
}

What does foo(“abcd”) evaluate to? It is is my understanding that this would reverse the input, but why is that?

Answer 1

You are facing recursion here!

Each level of the recursion is appending the first character to the end, and invoke a recursive call on the suffix of the string.

In your example, the call stack will look something like that:

s = "abcd" => append 'a' to the end, and invoke on "bcd".
s = "bcd" => append 'b' to the end, and invoke on "cd".
s = "cd" => append 'c' to the end, and invoke on "d".
s= = "d" => return "d" as it is.

when you go back from the recursion, you actually append it in reverse order:

return "d"
return "d" + 'c' (="dc")
return "dc" + 'b' (="dcb")
return "dcb" + 'a' (="dcba")

Answer 2

Hope this makes it clear how the recursion works in this case:

foo("bcd") + "a"
  (foo("cd") + "b") + "a"
    ((foo("d") + "c") + "b") + "a"
      (("d" + "c") + "b") + "a" -> "dcba"

Answer 3

It is a recursive reverse. s.substring(1) is the line without its first character; s.charAt(0) is the first character.

What the function is saying is "if the line is one character long, the answer is the line itself; otherwise, chop off the first character, compute the same function, and add the chopped off character to the end of the result".

You can work out on a piece of paper how performing the steps above amounts to reversing a string.

EDIT : It is worth noting that this implementation is going to crash with an exception if you try passing it an empty string. Changing if (s.length() == 1) to if (s.length() == 0) would address this problem (thanks to Tom Hawtin - tackline for mentioning this in a comment).