Leetcode——242. Valid Anagram

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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; }
};

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