Ransom Note

题目地址：https://leetcode.com/problems/ransom-note/?tab=Description

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

像堆积木一样，ransomNote相当于结果，magazine相当于原料，原料提供的决不能比结果所需原料的量要少，那我们可以分别对ransomNote的需要的“原料”与magazine提供的“原料”进行统计，然后做对比即可：

public class RansomNote {
    public static boolean canConstruct(String ransomNote, String magazine) {
        if (ransomNote.length() > magazine.length())
            return false;

        if (ransomNote.length() == 0)
            return true;

        Map<Character, Integer> ransomNodeMap = new HashMap<>();
        Map<Character, Integer> magazineMap = new HashMap<>();

        for (int i = 0; i < ransomNote.length(); i++) {
            if (!ransomNodeMap.containsKey(ransomNote.charAt(i))) {
                ransomNodeMap.put(ransomNote.charAt(i), 1);
            } else {
                ransomNodeMap.put(ransomNote.charAt(i), ransomNodeMap.get(ransomNote.charAt(i)) + 1);
            }
        }

        for (int i = 0; i < magazine.length(); i++) {
            if (!magazineMap.containsKey(magazine.charAt(i))) {
                magazineMap.put(magazine.charAt(i), 1);
            } else {
                magazineMap.put(magazine.charAt(i), magazineMap.get(magazine.charAt(i)) + 1);
            }
        }

        for (Map.Entry entry : ransomNodeMap.entrySet()) {
            if (!magazineMap.containsKey(entry.getKey()))
                return false;
            if ((int)entry.getValue() > magazineMap.get(entry.getKey()))
                return false;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(canConstruct("affbba", "aaaabbbbfff"));
    }
}

如果不算映射到map所花费的时间，那么这个方法的总时间复杂度略大于m+n。

虽然可行，但是效率略低，应该想一个更好的办法……