383. 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

链接：https://leetcode.com/problems/ransom-note/#/description

3/20/2017

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
        HashMap<Character, Integer> hs = new HashMap<Character, Integer>();
        
        for (int i = 0; i < magazine.length(); i++) {
            if (hm.containsKey(magazine.charAt(i))) {
                hm.put(magazine.charAt(i), hm.get(magazine.charAt(i)) + 1);
            } else {
                hm.put(magazine.charAt(i), 1);
            }
        }
        for (int i = 0; i < ransomNote.length(); i++) {
            if (hs.containsKey(ransomNote.charAt(i))) {
                hs.put(ransomNote.charAt(i), hs.get(ransomNote.charAt(i)) + 1);
            } else {
                hs.put(ransomNote.charAt(i), 1);
            }            
        }
        for (HashMap.Entry<Character, Integer> entry : hs.entrySet()) {
            Character key = entry.getKey();
            Integer value = entry.getValue();
            if (!hm.containsKey(key) || hm.get(key) < value) return false;
        }
        return true;
    }
}

别人不用hashmap只用array的解法：

https://discuss.leetcode.com/topic/53864/java-o-n-solution-easy-to-understand/10

Python的Collections.counter的解法：

def canConstruct(self, ransomNote, magazine):
    return not collections.Counter(ransomNote) - collections.Counter(magazine)