383. Ransom Note
2017-01-15 16:17
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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.
英文水平就是有限,第一遍没有读懂,以为是求一个字符串是否包含在另一个字符串中,直接用magazine.indexOf(ransomNote)!=-1分分钟就可以出结果了,在leetcode上运行才知道自己太年轻,毕竟leetcode上不会有这么白痴的问题...
public class fuzhi {
public boolean canConsrtruct(String ransomNote,String magazine){
int[] maga=new int[26];
int[] ran=new int[26];
for(int i=0;i<magazine.length();i++){
maga[magazine.charAt(i)-'a']++;
}
for(int i=0;i<ransomNote.length();i++){
ran[ransomNote.charAt(i)-'a']++;
}
for(int i=0;i<26;i++){
if(maga[i]<ran[i]){
return false;
}
}
return true;
}
public static void main(String[] args){
String ransomNote="abdhkash",magazine="aabcsakdhbs";
fuzhi r = new fuzhi();
System.out.println(r.canConsrtruct(ransomNote, magazine));
}
}
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
英文水平就是有限,第一遍没有读懂,以为是求一个字符串是否包含在另一个字符串中,直接用magazine.indexOf(ransomNote)!=-1分分钟就可以出结果了,在leetcode上运行才知道自己太年轻,毕竟leetcode上不会有这么白痴的问题...
题意:
给定一个字符串ransom和一个字符串magazine,判定ransom中的字符是否均在magazine中出现,且magazine中的字符每个只能使用一次。解法:
magazine中每个字符出现的次数必须大于等于ransom中字符出现的次数。因此newl两个数组来分别储存magazine、random中的次数,如果maga[i]<ran[i],返回false。public class fuzhi {
public boolean canConsrtruct(String ransomNote,String magazine){
int[] maga=new int[26];
int[] ran=new int[26];
for(int i=0;i<magazine.length();i++){
maga[magazine.charAt(i)-'a']++;
}
for(int i=0;i<ransomNote.length();i++){
ran[ransomNote.charAt(i)-'a']++;
}
for(int i=0;i<26;i++){
if(maga[i]<ran[i]){
return false;
}
}
return true;
}
public static void main(String[] args){
String ransomNote="abdhkash",magazine="aabcsakdhbs";
fuzhi r = new fuzhi();
System.out.println(r.canConsrtruct(ransomNote, magazine));
}
}
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