解法一:
class Solution {
public:
bool isAnagram(string s, string t) {
sort(s.begin(),s.end());
sort(t.begin(),t.end());
return s==t;
}
};
解法二:
非常巧妙的一种解法,避免了定义两个counts再进行比较,非常巧妙!!学习了!
class Solution {
public:
bool isAnagram(string s, string t) {
int counts[26];
memset(counts, 0, sizeof(counts));
if(s.size()!=t.size())
return false;
for (int i=0; i<s.size(); ++i) {
counts[s[i]-'a']++;
counts[t[i]-'a']--;
}
for (int i=0; i<26; ++i) {
if(counts[i]!=0)
return false;
}
return true;
}
};