给定两个字符串数组，求出二者不同的元素

题目描述：给定s = {"a","b","c","d","f","e","ab"}; s2 = {"a","ab"}; 求出s中不包含s2中的剩余元素。可以认为是 s-s2的结果。

思路： 利用hashmap，将s中的元素都存入hashmap，并将<K,V>中的value值赋值1，然后遍历s2中的元素，如果s1中存在则value+1，最后找出value为1的元素就是所求元素。

import java.util.*;
import java.util.Map.Entry;
public class StringMinute {
    public static void main(String[] args){
        String[] s = {"a","b","c","d","f","e","ab"};
        String[] s2 = {"a","ab"};
        
        Map<String, Integer> map = new HashMap<>();
        for (int i=0; i<s.length; i++){
            map.put(s[i], 1);
        }
        for (int i=0; i<s2.length; i++){
            if (map.containsKey(s2[i])){
                int count = map.get(s2[i]);
                count++;
                map.put(s2[i], count);
            }
        }
        Iterator iter = map.entrySet().iterator();
        while (iter.hasNext()){
            Map.Entry<String, Integer> entry = (Entry<String, Integer>)iter.next();
            if (entry.getValue() == 1){
                System.out.println(entry.getKey());
            } 
        }
    }
}