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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
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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; }
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原文地址:http://blog.csdn.net/yeqiuzs/article/details/52188263