思路：分别用两个map记录ransomNote和magazine中的字符以及出现的次数。最后遍历记录ransomNote的map，如果ransomNote的map中出现的magazine的map中没有出现或者出现的次数小于ransomNote的map则返回false，否则返回true；

class Solution {public boolean canConstruct(String ransomNote, String magazine) {Map<Character,Integer> ransomNote_map = new HashMap<>();Map<Character,Integer> magazine_map = new HashMap<>();for(int i = 0 ; i < ransomNote.length() ; i++){ransomNote_map.put(ransomNote.charAt(i), ransomNote_map.getOrDefault(ransomNote.charAt(i), 0) + 1);}for(int i = 0 ; i < magazine.length() ; i++){magazine_map.put(magazine.charAt(i),magazine_map.getOrDefault(magazine.charAt(i),0) + 1);}for(Character c : ransomNote_map.keySet()){if(!magazine_map.containsKey(c) || ransomNote_map.get(c) > magazine_map.get(c)){return false;}}return true;}
}