242. Effective letter ectopic words (implemented in java) --LeetCode

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topic

Given two strings s and t, write a function to determine whether t is an anagram of s.

Example 1:

输入: s = "anagram", t = "nagaram"
输出: true

Example 2:

输入: s = "rat", t = "car"
输出: false

Description:

You can assume that the string contains only lowercase letters.

Advanced:

What if the input string contains unicode characters? Can you adjust your solution to deal with this situation?

Solution 1: Sort

/**
 * 思路：
 * 简单过滤：看两个字符串长度是否一致
 * 转换成字符数组
 * 排序
 * 比较数组
 */
    public boolean isAnagram(String s, String t) {
    
    
        if (s.length()!=t.length())return false;
        char[] s1 = s.toCharArray();
        char[] t1 = t.toCharArray();

        Arrays.sort(s1);
        Arrays.sort(t1);

        return Arrays.equals(s1,t1);
    }

Time complexity: Onlogn

Space complexity: On
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Solution 2: Array

/**
 * 思路：
 * 简单过滤：看两个字符串长度是否一致
 * 数组长度26，代表26个字母
 * 遍历s和t，s中的字母在数组上++，t中的字母--
 * 最后遍历数组，如果有不是0的就不为有效字母异位词
 */
     public boolean isAnagram(String s, String t) {
    
    

        if (s.length()!=t.length())return false;

        int[] arr=new int[26];

        for (char c:s.toCharArray()){
    
    
            arr[c-'a']++;
        }

        for (char c:t.toCharArray()){
    
    
            arr[c-'a']--;
        }

        for (int i:arr)if (i!=0)return false;

        return true;
    }

Time complexity: On

Space complexity: O1
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Solution 3: HashMap

/**
 * 思路：
 * 简单过滤：看两个字符串长度是否一致
 * map中存放26个字母，以及出现的次数
 * s中出现++，t中出现--
 * 最后看map中的每个key的value是否为0
 */
     public boolean isAnagram(String s, String t) {
    
    
        if (s.length()!=t.length())return false;

        HashMap<Character, Integer> map = new HashMap<>();

        for (char c='a';c<='z';c++)map.put(c,0);

        for (int i=0;i<s.length();i++){
    
    
            map.put(s.charAt(i),map.get(s.charAt(i))+1);
            map.put(t.charAt(i),map.get(t.charAt(i))-1);
        }

        for (int i:map.values())if (i!=0)return false;

        return true;
    }

Time complexity: On

Space complexity: On
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Solution 4: HashMap (violence)

/**
 * 思路：
 * 将字符串转换为字符数组
 * 存储每个字符的次数在map中
 * 对比map中的数是否相同
 */
     public boolean isAnagram(String s, String t) {
    
    
        HashMap<Character, Integer> s_map = new HashMap<>();
        HashMap<Character, Integer> t_map = new HashMap<>();

        char[] s_chars = s.toCharArray();
        char[] t_chars = t.toCharArray();

        for (char c:s_chars){
    
    
            if (s_map.containsKey(c)) {
    
    
                s_map.put(c, s_map.get(c) + 1);
            }else {
    
    
                s_map.put(c, 1);
            }
        }

        for (char c:t_chars){
    
    
            if (t_map.containsKey(c)) {
    
    
                t_map.put(c, t_map.get(c) + 1);
            }else {
    
    
                t_map.put(c, 1);
            }
        }

        if (s_map.size()!=t_map.size())return false;

        for (Character c:s_map.keySet()){
    
    
            if (!t_map.containsKey(c)||!s_map.get(c).equals(t_map.get(c)))return false;
        }
        return true;
    }

Time complexity: On

Space complexity: On
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