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<Character, Integer> 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<Character, Integer> entry : strMap.entrySet()) {

if (entry.getValue() != 0) {

return false;

}

}

return true;

}

}

```