LeetCode: 49. Group Anagrams

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49. Group Anagrams

LeetCode - Group Anagrams

计数字符数量的解法有些精巧！这里想起来了波利亚在《怎样解题》中，着重强调的一点：一点要充分挖掘题目中的一直变量。题目中已经明确给出，所有的字符都是小写字符。但却没有意识到它的重要性！

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

All inputs will be in lowercase.
The order of your output does not matter.

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/**
 * Runtime: 16 ms, faster than 22.85% of Java online submissions for Group Anagrams.
 *
 * Memory Usage: 40.4 MB, less than 96.49% of Java online submissions for Group Anagrams.
 */
public List<List<String>> groupAnagrams(String[] strs) {
    if (Objects.isNull(strs) || strs.length == 0) {
        return Collections.emptyList();
    }
    Map<String, List<String>> charsToStringsMap = new HashMap<>();
    int[] count = new int[26];
    for (String str : strs) {
        Arrays.fill(count, 0);
        char[] chars = str.toCharArray();
        for (char aChar : chars) {
            count[aChar - 'a']++;
        }
        StringBuilder builder = new StringBuilder(26 * 2);
        for (int i : count) {
            builder.append("#").append(i);
        }
        String key = builder.toString();
        List<String> strings = charsToStringsMap.getOrDefault(key, new ArrayList<>());
        strings.add(str);
        charsToStringsMap.put(key, strings);
    }
    return new ArrayList<>(charsToStringsMap.values());
}

/**
 * Runtime: 8 ms, faster than 96.90% of Java online submissions for Group Anagrams.
 *
 * Memory Usage: 41.1 MB, less than 95.32% of Java online submissions for Group Anagrams.
 */
public List<List<String>> groupAnagramsSort(String[] strs) {
    if (Objects.isNull(strs) || strs.length == 0) {
        return Collections.emptyList();
    }
    Map<String, List<String>> charsToStringsMap = new HashMap<>();
    for (String str : strs) {
        char[] chars = str.toCharArray();
        Arrays.sort(chars);
        String key = new String(chars);
        List<String> strings = charsToStringsMap.getOrDefault(key, new ArrayList<>());
        strings.add(str);
        charsToStringsMap.put(key, strings);
    }
    return new ArrayList<>(charsToStringsMap.values());
}