[算法][力扣242]有效的字母异位词

big_rabbit05022026-03-14 9:17

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。

示例 1:

输入: s = "anagram", t = "nagaram"

输出: true

示例 2:

输入: s = "rat", t = "car"

输出: false

解法一：排序后对比

java 复制代码

public class Test_validAnagram {

    public static void main(String[] args) {
        String s = "anagram";
        String t = "nagaram";
        boolean ret = validAnagram(s, t);
        System.out.println(ret);
    }

    public static boolean validAnagram(String s1, String s2) {
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();

        Arrays.sort(chars1);
        Arrays.sort(chars2);

        return Arrays.equals(chars1, chars2);
    }
}

解法二：使用HashMap统计每次字符出现的次数

java 复制代码

public class Test_validAnagram2 {

    public static void main(String[] args) {
        String s = "anagram";
        String t = "nagaraam";
        boolean ret = validAnagram(s, t);
        System.out.println(ret);
    }

    public static boolean validAnagram(String s1, String s2) {
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();

        // 用 HashMap 来记录字符串 s1 中出现的字符的次数
        HashMap<Character, Integer> record = new HashMap<>();
        for (char ch : chars1) {
            record.put(ch, record.getOrDefault(ch, 0) + 1);
        }

        // 遍历char2，同事比较元素
        for (char ch2 : chars2) {
            int count = record.getOrDefault(ch2, 0) - 1;
            record.put(ch2, count);
            if (count < 0) {
                return false;
            }
        }
        return true;
    }
}

解法三：使用数组统计字符个数

java 复制代码

public class Test_validAnagram3 {

    public static void main(String[] args) {
        String s = "anagram";
        String t = "nagaram";
        boolean ret = validAnagram(s, t);
        System.out.println(ret);
    }

    public static boolean validAnagram(String s1, String s2) {
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();

        // 用 HashMap 来记录字符串 s1 中出现的字符的次数
        int[] record = new int[26];
        for (char ch : chars1) {
            record[ch - 'a']++;
        }

        // 遍历char2，同事比较元素
        for (char ch2 : chars2) {
            record[ch2 - 'a']--;
            if (record[ch2 - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
}