383. 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

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        if(ransomNote.length() == 0) return true;
        else if(magazine.length() == 0) return false;
        int[] arr = new int[26];
        
        for(int i = 0; i < magazine.length(); i++) arr[magazine.charAt(i) - 'a']++;
        int cou = 0;
        for(int i = 0; i < ransomNote.length(); i++){
            char c = ransomNote.charAt(i);
            if(arr[c - 'a'] > 0){
                --arr[c-'a'];
                cou++;
            }
            else return false;
        }
        if(cou == ransomNote.length()) return true;
        return false;
    }
}

问能否从magazine里组成ransomNote，把magazine先存到数组里，然后再遍历ransomNote，看是不是能组成就可以。

而且空字符也行，我佛辣

 

update20220824

class Solution {
    public boolean canConstruct(String r, String m) {
        int[] ch = new int[26];
        for(char c : m.toCharArray()) ch[c -'a']++;
        for(char c : r.toCharArray()) {
            ch[c - 'a']--;
            if(ch[c - 'a'] < 0) return false;
        }
        return true;
    }
}

不知道当时在纠结什么，用字典两次遍历就完事了