python3 解法, 执行用时: 72 ms, 内存消耗: 16.2 MB, 提交时间: 2023-10-28 12:47:16

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        record = [0] * 26
        for i in range(len(s)):
            #并不需要记住字符a的ASCII，只要求出一个相对数值就可以了
            record[ord(s[i]) - ord("a")] += 1
        print(record)
        for i in range(len(t)):
            record[ord(t[i]) - ord("a")] -= 1
        for i in range(26):
            if record[i] != 0:
                #record数组如果有的元素不为零0，说明字符串s和t 一定是谁多了字符或者谁少了字符。
                return False
                #如果有一个元素不为零，则可以判断字符串s和t不是字母异位词
                break
        return True

    # Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）：
    def isAnagram2(self, s: str, t: str) -> bool:
        from collections import defaultdict
        
        s_dict = defaultdict(int)
        t_dict = defaultdict(int)

        for x in s:
            s_dict[x] += 1
        
        for x in t:
            t_dict[x] += 1

        return s_dict == t_dict

python3 解法, 执行用时: 44 ms, 内存消耗: 16.3 MB, 提交时间: 2023-10-28 12:46:36

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        count = {}
        for char in s:
            if char in count:
                count[char] += 1
            else:
                count[char] = 1
        for char in t:
            if char in count:
                count[char] -= 1
            else:
                return False
        for value in count.values():
            if value != 0:
                return False
        return True

golang 解法, 执行用时: 8 ms, 内存消耗: 2.7 MB, 提交时间: 2023-10-28 12:45:45

func isAnagram1(s, t string) bool {
    var c1, c2 [26]int
    for _, ch := range s {
        c1[ch-'a']++
    }
    for _, ch := range t {
        c2[ch-'a']++
    }
    return c1 == c2
}

func isAnagram2(s, t string) bool {
    s1, s2 := []byte(s), []byte(t)
    sort.Slice(s1, func(i, j int) bool { return s1[i] < s1[j] })
    sort.Slice(s2, func(i, j int) bool { return s2[i] < s2[j] })
    return string(s1) == string(s2)
}

// with unicode
func isAnagram(s, t string) bool {
    if len(s) != len(t) {
        return false
    }
    cnt := map[rune]int{}
    for _, ch := range s {
        cnt[ch]++
    }
    for _, ch := range t {
        cnt[ch]--
        if cnt[ch] < 0 {
            return false
        }
    }
    return true
}

cpp 解法, 执行用时: 4 ms, 内存消耗: 7.5 MB, 提交时间: 2023-10-28 12:44:48

class Solution {
public:
    // sort
    bool isAnagram1(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t;
    }

    // hash
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) {
            return false;
        }
        vector<int> table(26, 0);
        for (auto& ch: s) {
            table[ch - 'a']++;
        }
        for (auto& ch: t) {
            table[ch - 'a']--;
            if (table[ch - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
};

java 解法, 执行用时: 15 ms, 内存消耗: 42.5 MB, 提交时间: 2023-10-28 12:43:56

class Solution {
    // sort
    public boolean isAnagram1(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        char[] str1 = s.toCharArray();
        char[] str2 = t.toCharArray();
        Arrays.sort(str1);
        Arrays.sort(str2);
        return Arrays.equals(str1, str2);
    }
    
    // hash
    public boolean isAnagram2(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        int[] table = new int[26];
        for (int i = 0; i < s.length(); i++) {
            table[s.charAt(i) - 'a']++;
        }
        for (int i = 0; i < t.length(); i++) {
            table[t.charAt(i) - 'a']--;
            if (table[t.charAt(i) - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
    
    // 输入字符串包含 unicode 字符
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        Map<Character, Integer> table = new HashMap<Character, Integer>();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            table.put(ch, table.getOrDefault(ch, 0) + 1);
        }
        for (int i = 0; i < t.length(); i++) {
            char ch = t.charAt(i);
            table.put(ch, table.getOrDefault(ch, 0) - 1);
            if (table.get(ch) < 0) {
                return false;
            }
        }
        return true;
    }
}

golang 解法, 执行用时: 0 ms, 内存消耗: 2.8 MB, 提交时间: 2021-07-18 08:32:03

func isAnagram(s string, t string) bool {
    
    if len(s) != len(t) {
        return false
    }
    cnt := [26]int{}
    for _, ch := range s {
        cnt[ch-'a']++
    }
    for _, ch := range t {
        cnt[ch-'a']--
    }
    for _, c := range cnt[:] {
        if c != 0 {
            return false
        }
    }
    return true
}

golang 解法, 执行用时: 12 ms, 内存消耗: 3 MB, 提交时间: 2021-06-12 03:28:54

func isAnagram(s string, t string) bool {
    if len(s) != len(t) {
        return false
    }
    b1, b2 := []byte(s), []byte(t)
    sort.Slice(b1, func(i, j int) bool {
        return b1[i] < b1[j]
    })
    sort.Slice(b2, func(i, j int) bool {
        return b2[i] < b2[j]
    })
    return string(b1) == string(b2)
}