Leetcode——242. Valid Anagram

题目

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

For example,
s = “anagram”, t = “nagaram”, return true.
s = “rat”, t = “car”, return false.

Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

解答

Anagram：n. 相同字母异序词，易位构词，变位词

My AC code

class Solution {//unordered_map<char,int>hash or int array[256]
public:
    bool isAnagram(string s, string t) {
        int m1[256]={-1},m2[256]={-1};
        if(s.length()!=t.length()) return false;
        for(int i=0;i<s.length();i++)
        {
            m1[s[i]]++;
            m2[t[i]]++;
        }
        for(int i=0;i<256;i++)
            if(m1[i]!=m2[i])
                return false;
        return true;
    }
};

Discuss AC code

hash method

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) return false;
        int n = s.length();
        unordered_map<char, int> counts;
        for (int i = 0; i < n; i++) {
            counts[s[i]]++;
            counts[t[i]]--;
        }
        for (auto count : counts)
            if (count.second) return false;
        return true;
    }
};

the improvement of my method

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.length() != t.length()) return false;
        int n = s.length();
        int counts[26] = {0};
        for (int i = 0; i < n; i++) { 
            counts[s[i] - 'a']++;
            counts[t[i] - 'a']--;
        }
        for (int i = 0; i < 26; i++)
            if (counts[i]) return false;
        return true;
    }
};

Another method
sorting

class Solution {
public:
    bool isAnagram(string s, string t) { 
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t; 
    }
};