题目链接:https://leetcode.com/problems/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.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
Subscribe to see which companies asked this question
思路:
题意是让你判断 前面字符串能否由后面字符串的部分字符构造出来。
因为只有小写字母,用数组代替map存储magazine每个字符出现的次数,遍历ransom note 每个字符并判断。
easy
算法:
public boolean canConstruct(String ransomNote, String magazine) {
int[] map = new int[26];
for(char mc:magazine.toCharArray()){
map[mc-'a']++;
}
for(char rc:ransomNote.toCharArray()){
if(map[rc-'a']>0){
map[rc-'a']--;
}else{
return false;
}
}
return true;
}原文:http://blog.csdn.net/yeqiuzs/article/details/52188263