242. Valid Anagram

發表於 | 分類於

problem

solution

option 1 - sorting

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

option 2 - hash table

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class Solution {
public:
    bool isAnagram(string s, string t) {
        vector<int> vec(26,0);
        for(char c:s) vec[c-'a']++;
        for(char c:t) vec[c-'a']--;
        return vec==vector<int>(26,0);
    }
};

analysis

  • option 1
    • time complexity O(n)
    • space complexity O(1)
  • option 2
    • time complexity O(n)
    • space complexity O(1)