解法1:排序,首先比较两个字符串长度,然后进行排序,作比较,
class Solution {
public:
bool isAnagram(string s, string t) {
if(s.length() != t.length())
{
return false;
}
sort(s.begin(),s.end());
sort(t.begin(),t.end());
if (s == t)
{
return true;
}else{
return false;
}
}
};
class Solution {
public:
bool isAnagram(string s, string t) {
if(s.length() != t.length())
{
return false;
}
sort(s.begin(),s.end());
sort(t.begin(),t.end());
if (s.compare(t) == 0)
{
return true;
}else{
return false;
}
}
};
解法2:哈希解法
class Solution {
public:
bool isAnagram(string s, string t) {
if (s.length() != t.length()) {
return false;
}
int record[26] = {0};
for (int i = 0; i < s.size(); i++)
{
record[s[i] - 'a']++;
}
for (int i = 0; i < s.size(); i++)
{
record[t[i] - 'a']--;
}
for (int i = 0; i < 26; i++)
{
if(record[i] != 0)
{
return false;
}
}
return true;
}
};