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    力扣链接：242. 有效的字母异位词，难度：简单。

    给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的 字母异位词。

    字母异位词是通过重新排列不同单词或短语的字母而形成的单词或短语，并使用所有原字母一次。

    示例 1:

    输入: s = "anagram", t = "nagaram"

    输出: true

    示例 2:

    输入: s = "rat", t = "car"

    输出: false

    约束:

    • 1 <= s.length, t.length <= 5 * 10^4
    • s 和 t 仅包含小写字母

    思路
    上一题下一题

    1. 如果两个字符串长度不同，直接返回false。
    2. 用两个Hash tables分别存储对两个字符串的统计数据，key为字符，value为该字符出现次数。
    3. 比较这两个Hash tables，看它们是否相等。

    复杂度

    时间复杂度

    O(N)

    空间复杂度

    O(N)

    题解语言： Java Python JavaScript C# Go Ruby C++

    Java # 

    class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        var sCharToCount = new HashMap<Character, Integer>();
        for (var character : s.toCharArray()) {
            sCharToCount.put(character, sCharToCount.getOrDefault(character, 0) + 1);
        }

        var tCharToCount = new HashMap<Character, Integer>();
        for (var character : t.toCharArray()) {
            tCharToCount.put(character, tCharToCount.getOrDefault(character, 0) + 1);
        }

        return sCharToCount.equals(tCharToCount);
    }
}

    Python # 

    # from collections import defaultdict

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False

        s_char_to_count = defaultdict(int)
        for char in s:
            s_char_to_count[char] += 1

        t_char_to_count = defaultdict(int)
        for char in t:
            t_char_to_count[char] += 1

        return s_char_to_count == t_char_to_count

    JavaScript # 

    var isAnagram = function (s, t) {
  if (s.length != t.length) {
    return false;
  }

  const sCharToCount = new Map()
  const tCharToCount = new Map()

  for (const char of s) {
    sCharToCount.set(char, (sCharToCount.get(char) || 0) + 1)
  }

  for (const char of t) {
    tCharToCount.set(char, (tCharToCount.get(char) || 0) + 1)
  }

  return _.isEqual(sCharToCount, tCharToCount)
};

    C# # 

    public class Solution
{
    public bool IsAnagram(string s, string t)
    {
        if (s.Length != t.Length)
            return false;

        var sCharToCount = new Dictionary<char, int>();
        var tCharToCount = new Dictionary<char, int>();

        foreach (char character in s)
            sCharToCount[character] = sCharToCount.GetValueOrDefault(character, 0) + 1;

        foreach (char character in t)
            tCharToCount[character] = tCharToCount.GetValueOrDefault(character, 0) + 1;

        foreach (var entry in sCharToCount)
        {
            if (entry.Value != tCharToCount.GetValueOrDefault(entry.Key))
            {
                return false;
            }
        }

        return true;
    }
}

    Go # 

    import "reflect"

func isAnagram(s string, t string) bool {
    if len(s) != len(t) {
        return false
    }

    // Create frequency maps for both strings
    sCharCount := make(map[rune]int)
    for _, char := range s {
        sCharCount[char]++
    }

    tCharCount := make(map[rune]int)
    for _, char := range t {
        tCharCount[char]++
    }

    return reflect.DeepEqual(sCharCount, tCharCount)
}

    Ruby # 

    def is_anagram(s, t)
  return false if s.length != t.length

  s_char_to_count = Hash.new(0)
  t_char_to_count = Hash.new(0)

  s.each_char do |char|
    s_char_to_count[char] += 1
  end

  t.each_char do |char|
    t_char_to_count[char] += 1
  end

  s_char_to_count == t_char_to_count
end

    C++ # 

    class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }

        unordered_map<char, int> s_char_to_count;
        for (char character : s) {
            s_char_to_count[character]++;
        }

        unordered_map<char, int> t_char_to_count;
        for (char character : t) {
            t_char_to_count[character]++;
        }

        return s_char_to_count == t_char_to_count;
    }
};
    题解语言： Java Python JavaScript C# Go Ruby C++

    其它语言

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