E
把#of occurrences 存进HashMap, 第一个string 做加法，第二个string做减法。最后看是否有不等于0的作判断。
```
public class Solution {
/*
* @param A: a string
* @param B: a string
* @return: a boolean
*/
public boolean Permutation(String A, String B) {
if (A == null || B == null || A.length() != B.length()) {
return false;
}
final Map strMap = new HashMap<>();
for (int i = 0; i < A.length(); i++) {
final char charA = A.charAt(i);
final char charB = B.charAt(i);
if (!strMap.containsKey(charA)) {
strMap.put(charA, 0);
}
strMap.put(charA, strMap.get(charA) + 1);
if (!strMap.containsKey(charB)) {
strMap.put(charB, 0);
}
strMap.put(charB, strMap.get(charB) - 1);
}
for (Map.Entry entry : strMap.entrySet()) {
if (entry.getValue() != 0) {
return false;
}
}
return true;
}
}
```