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……