Valid Anagram

题目如下：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.

思路如下：先判断两个String的长度，若长度不一致返回false；长度一致，对String进行判断，采用hashMap的形式，把string转换为char[],存入Map,key值为char,value为每个元素在数组中出现的次数，然后把另一个数组中的元素和Map中的元素进行比较，若存在Map中，且出现的次数为1，则删除Map中的此元素，若在Map中出现的次数大于1，则把其Value值减1，若char不存在Map中则返回false.

public class Solution {
public boolean isAnagram(String s, String t) {
int tlength = t.length();
int slength = s.length();
if(tlength == slength){
HashMap<Character,Integer> map = new HashMap<Character, Integer>();
char[] schar = s.toCharArray();
char[] tchar = t.toCharArray();
for(int i = 0 ;i < slength; i++){
if(map.containsKey(schar[i])){
int j = map.get(schar[i])+1;
map.put(schar[i], j);
}else{
map.put(schar[i], 1);
}
}

for(int i = 0 ; i < tlength; i++ ){
if(map.containsKey(tchar[i])){
if(map.get(tchar[i])==1){
map.remove(tchar[i]);
}else{
int j = map.get(tchar[i]);
map.put(tchar[i], j-1);
}
}else{
return false;
}
}

return true;
}
return false;
}
}