广义缩写的算法

Question

I have a question about how recursion stack works.

The problem I am stuck at now is Generalized Abbreviation from leetcode

The question states that

A word's generalized abbreviation can be constructed by taking any number of non-overlapping substrings and replacing them with their respective lengths. For example, "abcde" can be abbreviated into "a3e" ("bcd" turned into "3"), "1bcd1" ("a" and "e" both turned into "1"), and "23" ("ab" turned into "2" and "cde" turned into "3").

Given a string word, return a list of all the possible generalized abbreviations of word. Return the answer in any order.

Input: word = "word"
Output: ["4","3d","2r1","2rd","1o2","1o1d","1or1","1ord","w3","w2d","w1r1","w1rd","wo2","wo1d","wor1","word"]

Input: word = "a"
Output: ["1","a"]

Here is the code I am struggling with

/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
    public static List<String> generateAbbreviations(String word){
        List<String> ans = new ArrayList<String>();
        backtrack(ans, new StringBuilder(), word, 0, 0);
        return ans;
    }
 
    // i is the current position
    // k is the count of consecutive abbreviated characters
    private static void backtrack(List<String> ans, StringBuilder builder, String word, int i, int k){
        int len = builder.length(); // keep the length of builder 
        if(i == word.length()){
            if (k != 0) builder.append(k); // append the last k if non zero
            ans.add(builder.toString());
        } else {
            // the branch that word.charAt(i) is abbreviated
            backtrack(ans, builder, word, i + 1, k + 1);
 
            // the branch that word.charAt(i) is kept
            if (k != 0) builder.append(k);
            builder.append(word.charAt(i));
            backtrack(ans, builder, word, i + 1, 0);
        }
 
       builder.setLength(len); // reset builder to the original state
       System.out.println("Length of Stringbuilder : " + builder.length() + ",  Current element : " + builder.toString() + " , len : " + len);
    }
    public static void main(String[] args) {
    List<String> result = Ideone.generateAbbreviations("ab");
    System.out.println("Generalized abbreviation are: " + result);
  }
}

Stdout

Length of Stringbuilder : 1,  Current element : 2, len : 0
Length of Stringbuilder : 2,  Current element : 1b, len : 2
Length of Stringbuilder : 2,  Current element : 1b, len : 0
Length of Stringbuilder : 2,  Current element : a1, len : 1
Length of Stringbuilder : 2,  Current element : ab, len : 2
Length of Stringbuilder : 2,  Current element : ab, len : 1
Length of Stringbuilder : 1,  Current element : a, len : 0

Output

["2","1b","a1","ab"]

From the stdout line 2 to 3, How does 'len' become 0 from what recursion stack?

Answer 1

Stdout on leetcode:

Length of Stringbuilder : 0,  Current element :  , len : 0
Length of Stringbuilder : 2,  Current element : 1b , len : 2
Length of Stringbuilder : 0,  Current element :  , len : 0
Length of Stringbuilder : 1,  Current element : a , len : 1
Length of Stringbuilder : 2,  Current element : ab , len : 2
Length of Stringbuilder : 1,  Current element : a , len : 1
Length of Stringbuilder : 0,  Current element :  , len : 0

The function calls will be like this(I will denote them using (i, k, lenOfBuilder)). lenOfBuilder is the length calculated at the first line of each function call.

For input word = "ab"

1st call (0, 0, 0) -> called from the main method.

2nd call (1, 1, 0) -> 1st backtrack() called from else condition of 1st call.

3rd call (2, 2, 0) -> 1st backtrack() called from else condition of 2nd call.

Now base condition hits as (i == 2 and k != 0). So "2" added to the answer.
Return back to 2nd call.

4th call (2, 0, 2) -> 2nd backtrack() called from else condition of 2nd call. Before this call we have added "k" and charAt(i) to the builder ie "1b"

Now base condition hits as (i == 2). But K is 0. So "1b" added to the answer.
Return back to 2nd call.

Else condition of 2nd call ends.

Now following statements will be executed from the 2nd call of the function.

builder.setLength(len); // As the length of builder in 2nd call was 0. So this sets the length of builder to zero -> "0".
Sysout statement will print this updated length i.e 0

This is how your code has updated the length of the builder to 0. As at the very first line of your 2nd call you have calculated the original length that was 0. And after the end of else condition of your 2nd call. You updated the length to 0 and prints 0.