[题目] 

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.

You may assume that both strings contain only lowercase letters.

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

[题目解析] 首先明确题意，两个字符串，第一个字符串中的字母都是从第二个字符串中取得的。可以使用hashmap，将第一个字符串读取入hashmap,key是字符，value是对应该字符的个数。

然后遍历第二个字符串，对hashmap进行操作，依次对hashmap中含有的字符个数进行--，如果个数<0,则说明map中(即第一个字符中)含有的字符，在第二个字符串中没有覆盖到。代码如下。

public  boolean canConstruct(String ransomNote, String magazine)  {        Map charmap = new HashMap
     
      ();        for(int index = 0; index < magazine.length(); index++){            char tmp = magazine.charAt(index);            if(charmap.containsKey(tmp)){                int count = (int) charmap.get(tmp);                count++;                charmap.put(tmp, count);            }else{                charmap.put(tmp, 1);            }        }        for(int index = 0; index < ransomNote.length(); index++){            char tmp = ransomNote.charAt(index);            if(charmap.containsKey(tmp)){                int count = (int) charmap.get(tmp);                count--;                if(count < 0) return false;                charmap.put(tmp, count);            }else{                return false;            }        }        return true;}
     

        考虑到字符串中只有小写字母，我们可以优化下代码，整体思路不变。代码如下。

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