无需排序实现Java字母异位词分组：关键代码解析与调试困惑咨询

Hey there! Let's clear up the confusion around those two lines of code first, then walk through how you can debug this more effectively.

Breaking Down the Mysterious Lines #

1. arr[str.charAt(i)-'a']++; #

This line is counting the frequency of each lowercase letter in the current string. Here's the play-by-play:

• str.charAt(i) grabs the i-th character of the string (like 'e' from "eat").
• Subtracting 'a' converts the character to a 0-25 index: since lowercase letters have consecutive ASCII values, 'a' - 'a' = 0, 'b' - 'a' = 1, all the way to 'z' - 'a' = 25.
• The array arr has 26 slots (one for each letter), so we increment the slot matching the current character. For "eat", this sets arr[0] = 1 (for 'a'), arr[4] = 1 (for 'e'), arr[19] = 1 (for 't'), and leaves all other slots at 0.

2. String ns = new String(arr); #

This converts the character array into a string—but here's why your IDE shows weird commas/unprintable characters:

• The arr array holds integer values 0-25, which correspond to non-printable ASCII control characters (like null, start-of-header, etc.). These don't display as readable text, hence the strange output.
• The critical detail here: anagrams will produce identical arr arrays, so their converted ns strings will be identical too. This lets us use ns as a key in the HashMap to group all anagrams together.

Next Steps to Debug & Understand Better #

• Watch the arr array instead of ns: In your IDE, add a watch for the arr variable. You'll see a list of counts (e.g., [1,0,0,0,1,...1,...] for "eat") which makes the frequency logic obvious. You can also add a print statement to log it:

  System.out.println(Arrays.toString(arr));
  

• Simulate manually with small examples: Grab test strings like "eat", "tea", "tan", and walk through how arr gets populated for each. You'll quickly see that anagrams have identical arr values—this is the core of the grouping logic.
• Replace ns with a human-readable key: If the unprintable string is throwing you off, modify the code to create a readable key (e.g., "a1e1t1" for "eat"). This won't change the algorithm's logic, just make debugging easier:

  StringBuilder sb = new StringBuilder();
  for (int i = 0; i < 26; i++) {
      if (arr[i] > 0) {
          sb.append((char)('a' + i)).append(arr[i]);
      }
  }
  String ns = sb.toString();
  

• Log the HashMap contents: Add System.out.println(map); after processing all strings. This will show you how each key maps to a list of anagrams, confirming the grouping works as expected.

内容的提问来源于stack exchange，提问作者rickygrimes