> 很简单的字符串查找比较题目 - Approach 1：Brute Force > 暴力求解，将字符串读取到数组中使用，然后挨个进行对比 - Approach 2 : set > 使用set数组进行字符的查找，然后判断是否相同。相比方法一，性能有一些提升 - Approach 3：array > 由于题目声明只是用小写字母，因此可以使用array代替set实现 - Approach 4：sort > 使用sort对两个字符串进行排序，思路简单但是实现很耗时

array.cpp

//Runtime: 12 ms, faster than 97.85%
//Memory Usage: 9.2 MB, less than 58.97%

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

brute_force.cpp

//Runtime: 28 ms, faster than 25.25% 
//Memory Usage: 9.8 MB, less than 5.16%

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.length() != t.length()) return false;
        
        vector<char> array1,array2;
        for(int i = 0;i < s.length();++i){
            array1.push_back(s[i]);
            array2.push_back(t[i]);
        }
        sort(array1.begin(),array1.end());
        sort(array2.begin(),array2.end());
        
        for(int i = 0;i < s.length();++i)
            if(array1[i] != array2[i])
                return false;
        return true;
    }
};

set.cpp

//Runtime: 16 ms, faster than 58.39% 
//Memory Usage: 9.3 MB, less than 15.23%

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.length() != t.length()) return false;
        unordered_map<char,int> set1,set2;
        
        for(int i = 0;i < s.length();++i){
            set1[s[i]]++,set2[t[i]]++;
        }
        for(int i = 0;i < s.length();++i)
            if(set1[s[i]] != set2[s[i]])
                return false;
        return true;
    }
};

//进一步优化
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;
    }
};

sort.cpp

//Runtime: 28 ms, faster than 25.25%
//Memory Usage: 9.2 MB, less than 58.97%

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