*LeetCode--Ransom Note

Ransom Note   

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

看到这道题目的时候，第一时间想到的是map来进行记录每个单词的数量，然后进行判断

但是看讨论区后，发现可以直接用数组来进行记录，很方便的思路啊！！！

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        if(ransomNote == null || ransomNote.length() == 0) return true;
        if(magazine == null || magazine.length() == 0) return false;
        
        char[] note = ransomNote.toCharArray();
        char[] letter = magazine.toCharArray();
        
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i < letter.length; i++){
            map.put(letter[i], map.getOrDefault(letter[i], 0) + 1);
        }
        for(int i = 0; i < note.length; i++){
            Integer num = map.get(note[i]);
            if(num != null && num >= 1){
                map.put(note[i], num - 1);
            } else{
                return false;
            }
        }
        return true;
    }
}

　　

数组的方式：

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        if(ransomNote == null || ransomNote.length() == 0) return true;
        if(magazine == null || magazine.length() == 0) return false;
        
        int[] letter = new int[26];
        for(int i = 0; i < magazine.length(); i++){
            letter[magazine.charAt(i) - 'a']++;
        }
        for(int i = 0; i < ransomNote.length(); i++){
            if(--letter[ransomNote.charAt(i) - 'a'] < 0) return false;
        }
        return true;
    }
}