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[leetcode]ransom note

admin 11月 12, 2020 0

题目描述

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.

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canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

解题思路与代码

给定两个字符串magazine和ransomNote，问ransomNote是否可以从magazine中抽取字母（每个字母只能用一次）组成。思路挺简单，只要判断ransomNote中的字符是否全部在magazine中即可，利用hash即可。


public class  {
    public boolean canConstruct1(String ransomNote, String magazine) {
        
        int[] map = new int[26];

        // for(int i = 0; i < magazine.length(); i++){
        //     map[magazine.charAt(i) - 'a']++;
        // }

        // for(int i = 0; i < ransomNote.length(); i++){
        //     if(--map[ransomNote.charAt(i) - 'a'] < 0) return false;
        // }

        for(char ch : magazine.toCharArray()){
            map[ch - 'a']++;
        }

        for(char ch : ransomNote.toCharArray()){
            if(--map[ch - 'a'] < 0) return false;
        }

        return true;
    }

    public boolean canConstruct(String ransomNote, String magazine) {
        //可以用HashMap<Character, Integer>
        Map<Character, Integer> map = new HashMap<>();

        for(char ch : magazine.toCharArray()){
            if(map.containsKey(ch)){
                map.put(ch, map.get(ch) + 1);
            }else{
                map.put(ch, 1);
            }
        }

        for(char ch : ransomNote.toCharArray()){
            if(map.containsKey(ch)){
                int newCount = map.get(ch) - 1;
                if(newCount < 0) return false;
                map.put(ch, newCount);
            }else{
               return false;
            }
        }

        return true;
    }
}