[LeetCode] NO.383 Ransom Note

[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") -> false
canConstruct("aa", "ab") -> false
canConstruct("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<Character,Integer>();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;}

转载于:https://www.cnblogs.com/zzchit/p/5770355.html

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