javascript - 一個集合的元素均分給另一個數組物件, 有什麼方法比較簡單?

Question

1. 陣列A=["G","D","B","H"] , 元素個數不定, 元素內容可以是任意字元

2. #集合B=[{"id":"a",item:""},{"id":"a=b",item:""}], 數量不定, 結構固定

3. 現在要把A中的元素均分給B中的item,分到多個時用逗號分隔.

4. 對於A的元素個數小於或大於B的長度時, 只要求A要全部在B裡出現就行了, 集合B的item至少要分到一個元素, 最好均分, 但集合B裡的每個item不能有重複元素

#有什麼方法比較簡單?

Answer 1

已經解決了

import com.google.common.collect.Lists;
import com.google.common.collect.Maps;
import com.google.common.collect.Sets;
import org.apache.commons.collections.CollectionUtils;
import org.apache.commons.lang3.RandomStringUtils;
import org.apache.commons.lang3.RandomUtils;
import org.apache.commons.lang3.StringUtils;

import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Set;

/**
 * Created by YSTYLE on 2017-04-17 0017.
 */
public class TEst {

    private static List<String> list = Lists.newArrayList();
    private static List<Map> los = Lists.newArrayList();

    public static void main(String[] args) {
        init();
        List<String> augmented = list;
        int groupCount = los.size();
        if (list.size() < groupCount){
            augmented = Augmented(list, groupCount);
        }
        List<List<String>> chunk = chunk2(augmented, groupCount);
        for (int i = 0; i < los.size(); i++) {
            los.get(i).put("item", StringUtils.join(chunk.get(i),",") );
        }
        System.out.println(los);
    }
    // 初始化测试数据
    private static void init (){
        int losCount = RandomUtils.nextInt(1,10);
        int listCount = RandomUtils.nextInt(1,10);
        for (int i = 0; i < listCount ; i++) {
            list.add(RandomStringUtils.randomAlphabetic(4));
        }
        for (int i = 0; i <  losCount; i++) {
            Map<String,Integer> map = new HashMap<String, Integer>();
            map.put("id",RandomUtils.nextInt(10000,99999));
            los.add(map);
        }
        System.out.println("生成的数组: " + list+"  数量: "+listCount);
        System.out.println("生成的对象数量: " + los.size());
    }

    // 分组数据
    public static <T> List<List<T>> chunk2(List<T> list, int group){
        if (CollectionUtils.isEmpty(list)){
            return Lists.newArrayList();
        }
        List<List<T>> result = Lists.newArrayList();
        Map<Integer,Set<T>> temp = Maps.newHashMap();
        for (int i = 0; i < list.size(); i++) {
            if (temp.containsKey(i%group)) {
                Set<T> ts = temp.get(i % group);
                ts.add(list.get(i));
                temp.put(i%group,ts);
            }else {
                Set<T> ts = Sets.newHashSet();
                ts.add(list.get(i));
                temp.put(i % group,ts);
            }
        }
        for (Set<T> ts : temp.values()) {
            result.add(Lists.newArrayList(ts));
        }
        return result;
    }

    // 填充数据
    public static <T> List<T> Augmented(List<T> list ,int size){
        int length = CollectionUtils.isEmpty(list)?0:list.size();
        if (length<1){
            return Lists.newArrayList();
        }
        List<T> result = Lists.newArrayList(list);
        if (length > size){
            return result;
        }
        int count = size - length;
        for (int i = 0; i < count; i++) {
            result.add(list.get(RandomUtils.nextInt(0, length)));
        }
        return result;
    }
}

Answer 2

雷雷