Algorithms—242.Valid Anagram

思路：等价于判定两个字符串字母组成是否相同。

public class Solution {    public boolean isAnagram(String s, String t) {        if (s.length()!=t.length()) {return false;}        Map<Character,Integer> smap=new HashMap<Character, Integer>();        Map<Character,Integer> tmap=new HashMap<Character, Integer>();        char[] cs=s.toCharArray();        char[] ct=t.toCharArray();        for (int i = 0; i < ct.length; i++) {if (smap.get(cs[i])!=null) {smap.put(cs[i], smap.get(cs[i])+1);}else {smap.put(cs[i], 1);}if (tmap.get(ct[i])!=null) {tmap.put(ct[i], tmap.get(ct[i])+1);}else {tmap.put(ct[i], 1);}}        for (int i = 0; i < ct.length; i++) {if (!tmap.get(ct[i]).equals(smap.get(ct[i]))) {return false;}}        return true;    }}

耗时：448ms。