[LeetCode] 383. Ransom Note

轉自LeetCode

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

<Solution>
想法如下

• 使用一個 hash map 來記錄 magazine 出現的字元，以及次數

• 檢查在 ransom note 裡面的字元，是不是存在 hash map 中，以及沒有超過次數

code 如下

C++

	class Solution {
	public:
	bool canConstruct(string ransomNote, string magazine) {
	unordered_map<char, int> hashmap;
	for(const auto &c : magazine) {
	hashmap[c]++;
	}
	for(const auto &c : ransomNote) {
	if(!hashmap.count(c) || hashmap[c]-- < 1) {
	return false;
	}
	}
	return true;
	}
	};

view raw ransomNote.cpp hosted with ❤ by GitHub

Java

	class Solution {
	public boolean canConstruct(String ransomNote, String magazine) {
	HashMap<Character, Integer> map = new HashMap<>();
	char c;
	final int magaLen = magazine.length();
	for(int i = 0; i < magaLen; i++) {
	c = magazine.charAt(i);
	map.put(c, map.getOrDefault(c, 0) + 1);
	}
	final int ranLen = ransomNote.length();
	for(int i = 0; i < ranLen; i++) {
	c = ransomNote.charAt(i);
	if(!map.containsKey(c) || map.put(c, map.get(c)-1) < 1) {
	return false;
	}
	}
	return true;
	}
	}

view raw ransomNote.java hosted with ❤ by GitHub