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?
解答思路：
设置一个大小为26的整整型数组来记录s这个字符串中每个字母出现的次数，判断t字符串中每个字母出现的次数是否与之相同，若相同，则说明构成变位词。
代码如下：
class Solution {
public:
bool isAnagram(string s, string t) {
int s_size = s.length();
int t_size = t.length();
if(s_size != t_size)
{
return false;
}
if(s_size == 0)
return true;
char s_copy[s_size + 1];
char t_copy[t_size + 1];
strncpy(s_copy,s.c_str(),s_size);
strncpy(t_copy,t.c_str(),t_size);
int num[26]={0};
for(int i = 0;i < s_size;i++)
{
num[s_copy[i]-'a']++;
}
for(int i = 0;i < t_size;i++)
{
num[t_copy[i]-'a']--;
if(num[t_copy[i]-'a'] < 0)
return false;
}
for(int i = 0;i < 26;i++)
{
if(num[i] != 0)
return false;
}
return true;
}
};
、)
