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【Leetcode】Ransom Note

时间:2016-08-12 07:54:29      阅读:122      评论:0      收藏:0      [点我收藏+]

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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;
        
    }


【Leetcode】Ransom Note

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

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