2021-02-11 49. Group Anagrams

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该博客介绍了如何利用哈希映射解决编程问题，具体是将输入的一组字符串根据其字母异位词关系进行分组。通过计算每个字符串的字母素数乘积作为哈希键，有效地将相同字母异位词归为一类。这种方法适用于中等难度的编程挑战，涉及到的数据结构包括哈希表和列表，涉及的编程语言为Java。

49. Group Anagrams

Difficulty: Medium

Related Topics: Hash Table, String

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

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

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Constraints:

1 <= strs.length <= 10<sup>4</sup>
0 <= strs[i].length <= 100
strs[i] consists of lower-case English letters.

Solution

Language: Java

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        int[] a = new int[127];
        a['a']=2; a['b']=3; a['c']=5; 
        a['d']=7; a['e']=11; a['f']=13; 
        a['g']=17; a['h']=19; a['i']=23; 
        a['j']=29; a['k']=31; a['l']=37; 
        a['m']=41; a['n']=43; a['o']=47;
        a['p']=53; a['q']=59; a['r']=61; a['s']=67;
        a['t']=71; a['u']=73; a['v']=79;
        a['w']=83; a['x']=89; a['y']=97; a['z']=101;
        
        int[] nums = new int[strs.length];
        
        HashMap<Long, List<String>> hashmap = new HashMap<>();
        
        for (int i = 0; i < strs.length; i++) {
            long result = 2;
            for (int j = 0; j < strs[i].length(); j++) {
                result *= a[strs[i].charAt(j)];
            }
            if (hashmap.containsKey(result)) {
                List<String> list = hashmap.get(result);
                list.add(strs[i]);
                hashmap.put(result, list);
            } else {
                List<String> list = new ArrayList<>();
                list.add(strs[i]);
                hashmap.put(result, list);
            }
        }
        
        List<List<String>> solut = new ArrayList<>();
        
        for (List<String> value : hashmap.values()) {
            solut.add(value);
        }
        
        return solut;
        
        
    }
}