leetcode 383. Ransom Note

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:

You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

这个题嘛，很简单。
package leetcode;

public class Ransom_Note_383 {

public boolean canConstruct(String ransomNote, String magazine) {
int[] magazineChars=new int[26];
char[] chars1=ransomNote.toCharArray();
char[] chars2=magazine.toCharArray();
for(int i=0;i<chars2.length;i++){
int index=chars2[i]-'a';
magazineChars[index]++;
}
for(int i=0;i<chars1.length;i++){
int index=chars1[i]-'a';
magazineChars[index]--;
if(magazineChars[index]<0){
return false;
}
}
return true;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Ransom_Note_383 r=new Ransom_Note_383();
System.out.println(r.canConstruct("aa", "aab"));
}

}我的解法beat了98%的java submissions.
大神有解法跟我一样的，也有用map的。

public boolean canConstruct(String ransomNote, String magazine) {
Map<Character, Integer> magM = new HashMap<>();
for (char c:magazine.toCharArray()){
int newCount = magM.getOrDefault(c, 0)+1;
magM.put(c, newCount);
}
for (char c:ransomNote.toCharArray()){
int newCount = magM.getOrDefault(c,0)-1;
if (newCount<0)
return false;
magM.put(c, newCount);
}
return true;
}