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;     }};