Leetcode - 49 - Group Anagrams (medium)

字符串变位词分组问题的解决方案

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博客围绕字符串变位词分组问题展开，变位词是通过重新排列不同单词或短语字母形成的。给出两种解决方案，一是以排序后的数组为键创建映射，时间复杂度为O(nklogk)；二是按26个字符的计数分类，以计数为键，时间复杂度为O(NK)，并给出Java和Python示例。

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.

Solutions:

(1) sort equal, create a map with sorted array as key (ans.put(key, new ArrayList()), ans.get(key).add(s))

Time: O(nklogk) sort(O(klogk)) n times

Java:

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs.length == 0) return new ArrayList();
        Map<String, List> ans = new HashMap<String, List>();
        for (String s : strs) {
            char[] ca = s.toCharArray();
            Arrays.sort(ca);
            String key = String.valueOf(ca);
            if (!ans.containsKey(key)) ans.put(key, new ArrayList());
            ans.get(key).add(s);
        }
        return new ArrayList(ans.values());
    }
}

// https://leetcode.com/problems/group-anagrams/solution/

Python: collections.defaultdict

class Solution(object):
    def groupAnagrams(self, strs):
        ans = collections.defaultdict(list)
        for s in strs:
            ans[tuple(sorted(s))].append(s)
        return ans.values()

# https://leetcode.com/problems/group-anagrams/solution/

(2) categorize by the count of 26 charaters

use the count as the key of the map

Time : O(NK), where N is the length of strs, and K is the maximum length of a string in strs

Java:

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs.length == 0) return new ArrayList();
        Map<String, List> ans = new HashMap<String, List>();
        int[] count = new int[26];
        for (String s : strs) {
            Arrays.fill(count, 0);
            for (char c : s.toCharArray()) count[c - 'a']++;

            StringBuilder sb = new StringBuilder("");
            for (int i = 0; i < 26; i++) {
                sb.append('#');
                sb.append(count[i]);
            }
            String key = sb.toString();
            if (!ans.containsKey(key)) ans.put(key, new ArrayList());
            ans.get(key).add(s);
        }
        return new ArrayList(ans.values());
    }
}

// https://leetcode.com/problems/group-anagrams/solution/

Python:

class Solution:
    def groupAnagrams(strs):
        ans = collections.defaultdict(list)
        for s in strs:
            count = [0] * 26
            for c in s:
                count[ord(c) - ord('a')] += 1
            ans[tuple(count)].append(s)
        return ans.values()

# https://leetcode.com/problems/group-anagrams/solution/