LeetCode 383. Ransom Note

原题链接在这里：https://leetcode.com/problems/ransom-note/

题目：

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

题解：

第二个string的char们能否组成第一个string. 用freq来maintain magazine的么个char的个数，再把ransomNote的对应char个数减掉，若出现负数就不能通过magazine create ransomNote.

Time Complexity: O(magazine.length() + ransomNote.length()). Space: O(1)

AC Java:

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        if(ransomNote == null || magazine == null){
            throw new IllegalArgumentException("Invalid input string.");
        }
        
        int [] freq = new int[256];
        for(int i = 0; i<magazine.length(); i++){
            freq[magazine.charAt(i)]++;
        }
        
        for(int i = 0; i<ransomNote.length(); i++){
            if(--freq[ransomNote.charAt(i)] < 0){
                return false;
            }
        }
        
        return true;
    }
}

AC C++:

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int map[26] = {0};
        for(char c : magazine){
            map[c - 'a']++;
        }

        for(char c : ransomNote){
            if(map[c - 'a']-- == 0){
                return false;
            }
        }

        return true;
    }
};

跟上Stickers to Spell Word.