feat: solve group anagrams

group-anagrams/GangBean.java

@@ -0,0 +1,37 @@
+class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        /**
+        1. understanding
+        - grouping the anagrams together, and return groups
+        2. strategy
+        - anagram group's identity: same characters with same counts
+        - so, transform each strs to character and count hashtable, called 'id'.
+        - if groups contains strs's 'id', then append
+        - return values list
+        3. complexity
+        - time: O(N * L) where, N is the length of array strs, and L is the max length of each str
+        - space: O(N * L)
+        */
+        Map<Map<Character, Integer>, List<String>> groups = new HashMap<>();
+        for (String word: strs) {
+            Map<Character, Integer> id = idOf(word);
+            List<String> group = groups.getOrDefault(id, new ArrayList<>());
+            group.add(word);
+            groups.put(id, group);
+        }
+
+        // System.out.println(groups);
+        List<List<String>> ret = new ArrayList<>();
+        ret.addAll(groups.values());
+        return ret;
+    }
+
+    private Map<Character, Integer> idOf(String word) {
+        Map<Character, Integer> id = new HashMap<>();
+        for (char c: word.toCharArray()) {
+            id.put(c, id.getOrDefault(c, 0) + 1);
+        }
+        return id;
+    }
+}
+