[Leetcode] Ransom Note

来源:互联网 发布:马踏棋盘深度优先算法 编辑:程序博客网 时间:2024/06/02 11:33


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") -> falsecanConstruct("aa", "ab") -> falsecanConstruct("aa", "aab") -> true

都是套路,遇到字符个数,先申请数组,int[] cha=new int[26]

public class Solution {    public boolean canConstruct(String ransomNote, String magazine) {    int[] cha=new int[26];      for(int i=0;i<magazine.length();i++)      {          cha[magazine.charAt(i)-'a']+=1;      }      for(int i=0;i<ransomNote.length();i++)      {          cha[ransomNote.charAt(i)-'a']-=1;        if(cha[ransomNote.charAt(i)-'a']<0) return false;      }      return true;            }}


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