[leetcode] 7. valid anagram · effective letter ectopic words

Title Description

Description in English

Given two strings s and t, return true if t is an anagram of s, and false otherwise. 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.

English address

https://leetcode.com/problems/valid-anagram/

Description of Chinese version

Given two strings s and t , Write a function to determine t Whether it is s Letter heteronym of . Be careful ： if s and t Each character in the has the same number of occurrences , said s and t They are mutually alphabetic words .

Example 1: Input : s = "anagram", t = "nagaram" Output : true Example 2: Input : s = "rat", t = "car" Output : false

Advanced : If the input string contains unicode What about characters ？ Can you adjust your solution to deal with this situation ？

Constraints · Tips ：

• 1 <= s.length, t.length <= 5 * 104

• s and t Only lowercase letters

Address of Chinese version

https://leetcode.cn/problems/valid-anagram/

Their thinking

Use Map The structure stores the letters read and the number of occurrences , namely Map< character , Number of occurrences >

Specific steps ：

1. First compare the length of two strings , Inequality returns directly false

2. Same degree , At the same time through , From a string s Add the characters read in map（ If none, add directly , If there is one, it should key Times of value +1）, From a string t Characters read in move out map（ If nothing, add , The number of occurrences is -1, If there is one, it should key Times of value -1）

3. The number of times of traversal judgment removal is 0 Of key value

4. Judge map Is it empty ： Empty return true; Instead, return to false

How to solve the problem

My version

class Solution {
 public boolean isAnagram(String s, String t) {
        Map<Character, Integer> map = new HashMap<>();
        StringBuffer stringBuffer1 = new StringBuffer(s);
        int length1 = s.length();
        StringBuffer stringBuffer2 = new StringBuffer(t);
        int length2 = t.length();
        if (length1 != length2) {
            return false;
        }
        int len = length1;
        for (int i = 0; i < len; i++) {
            char cc = stringBuffer1.charAt(i);
            char tc = stringBuffer2.charAt(i);
            if (map.containsKey(cc)) {
                map.put(cc, map.get(cc) + 1);
            } else {
                map.put(cc, 1);
            }
            if (map.containsKey(tc)) {
                map.put(tc, map.get(tc) - 1);
            } else {
                map.put(tc, -1);
            }
            if (map.get(cc) != null) {
                if (map.get(cc) == 0) {
                    map.remove(cc);
                }
            }
            if (map.get(tc) != null) {
                if (map.get(tc) == 0) {
                    map.remove(tc);
                }
            }
        }

        if (map.isEmpty()) {
            return true;
        }
        return false;
    }
}

Official edition

Method 1 ： Sort

class Solution {
    public boolean isAnagram(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);
    }

}

Complexity analysis

• Time complexity ：O(nlogn)
• Spatial complexity ：O(logn)

Method 2 ： Hashtable

class Solution {
    public boolean isAnagram(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;
    }
}

Advanced questions

class Solution {
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;
    }
}

Complexity analysis

• Time complexity ：O(n), among n by s The length of
• Spatial complexity ：O(s), among s Is the character set size , here s=26

High praise version

​

class Solution {
 public boolean isAnagram(String s, String t) {
        if(s.length() != t.length())
            return false;
        int[] alpha = new int[26];
        for(int i = 0; i< s.length(); i++) {
            alpha[s.charAt(i) - 'a'] ++;
            alpha[t.charAt(i) - 'a'] --;
        }
        for(int i=0;i<26;i++)
            if(alpha[i] != 0)
                return false;
        return true;
    }

}

summary

The high praise answer is to use a set to store the order of characters , Plus... Appears in a string , If it appears in another string, it will decrease , The last occurrence times are 0 Then it proves that the two strings are exactly the same , But it is very clever that Gao Zan's reply is cleverly used 26 The order of letters and characters ASCII Code value , Only one size is used 26 Of int Arrays are cleverly stored 26 The order in which characters appear （ Unlike my clumsy use of key value pairs ）, But if we consider the advanced problem , Using key value pairs still has advantages .