LeetCode 383 Ransom Note

Posted 2020-08-06

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

 

思路：

前者对应字符的出现次数要小于等于后者对应字符出现的次数，这样即可判别为true。可以用两个list（或者一个queue一个list，一个stack一个list，均可），将前者的字符串一个个从所属数据结构中剔除时，在后者的数据结构中找到对应字符并剔除，如果无法完成该操作，则视为false。

 

解法：

 1 import java.util.ArrayList;
 2 import java.util.Stack;
 3 
 4 public class Solution
 5 {
 6     public boolean canConstruct(String ransomNote, String magazine)
 7     {
 8         Stack<Character> stack = new Stack<>();
 9         ArrayList<Character> list = new ArrayList<>();
10 
11         for(char c: ransomNote.toCharArray())
12             stack.push(c);
13         for(char c: magazine.toCharArray())
14             list.add(c);
15 
16         while(!stack.isEmpty())
17         {
18             if(list.contains(stack.peek()))
19                 list.remove(stack.pop());
20             else
21                 return false;
22         }
23 
24         return true;
25     }
26 }