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383. Ransom Note

383. Ransom Note

作者: namelessEcho | 来源:发表于2017-09-21 20:48 被阅读0次

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.

手写以ch-'a'建立的map可以更快,不过懒得写了。

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        HashMap<Character,Integer> map = new HashMap<>();
        for(int i = 0 ;i<magazine.length();i++)
        {
            char ch = magazine.charAt(i);
            map.put(ch,map.containsKey(ch)?map.get(ch)+1:1);
        }
        for(int i = 0 ;i<ransomNote.length();i++)
        {
            char ch = ransomNote.charAt(i);
            if(map.containsKey(ch))
            {
                int count =map.get(ch);
                if (count==0)
                    return false;
                else
                    map.put(ch,count-1);
            }
            else
                return false;
        }
        return true;
    }
}

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