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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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