Java判断两个ListMap值是否相等（包括数据长度、存储的值）超实用工具类（亲测）

╰+哭是因爲堅強的太久メ 2021-09-07 06:21 312阅读 0赞

## 一、功能描述 ##

比较两个list中存储的map数据，比较的list的数据类型需一致，就能比较出存储的数据是否一致。

## 二、实现步骤 ##

1、比较两个list的长度大小

2、统计出每个list中存储的map值的个数及重复的，进行比较

3、具体比较值

## 三、代码实例 ##

private static final Integer INTEGER_ONE = 1;
    
    // 判断并返回结果 true和false
    public static boolean isEqualCollection(Collection a, Collection b) {
        if (a.size() != b.size()) { // size是最简单的相等条件
            return false;
        }
        Map mapa = getCardinalityMap(a);
        System.out.println(mapa);
        Map mapb = getCardinalityMap(b);
        System.out.println(mapb);
    
        // 转换map后，能去掉重复的，这时候size就是非重复项，也是先决条件
        if (mapa.size() != mapb.size()) {
            System.out.println("存储的map数据不一致！");
            return false;
        } else {
            System.out.println("转换map后，能去掉重复的，这时候size就是非重复项后，存储的map数据一致！");
        }
        Iterator it = mapa.keySet().iterator();
        while (it.hasNext()) {
            Object obj = it.next();
            // 查询同一个obj，首先两边都要有，而且还要校验重复个数，就是map.value
            if (getFreq(obj, mapa) != getFreq(obj, mapb)) {
                return false;
            }
        }
        return true;
    }
    
    /**
     * 以obj为key，可以防止重复，如果重复就value++ 这样实际上记录了元素以及出现的次数
     */
    public static Map getCardinalityMap(Collection coll) {
        Map count = new HashMap();
        for (Iterator it = coll.iterator(); it.hasNext();) {
            Object obj = it.next();
            Integer c = (Integer) count.get(obj);
            if (c == null)
                count.put(obj, INTEGER_ONE);
            else {
                count.put(obj, new Integer(c.intValue() + 1));
            }
        }
        return count;
    }
    
    private static final int getFreq(Object obj, Map freqMap) {
        Integer count = (Integer) freqMap.get(obj);
        if (count != null) {
            return count.intValue();
        }
        return 0;
    }
    
    
    // main方法测试类
    public static void main(String[] args) {
        Map<String, Object> m1 = new HashMap<String, Object>();
        m1.put("a", "abc");
        m1.put("b", "123");
        m1.put("c", "123");
        m1.put("d", "123");
    
        Map<String, Object> m2 = new HashMap<String, Object>();
        m2.put("c", "123");
        m2.put("a", "abc");
        m2.put("b", "123");
        m2.put("d", "123");
    
        Map<String, Object> m3 = new HashMap<String, Object>();
        m3.put("a", "abc");
        m3.put("b", "123");
        m3.put("c", "123");
        m3.put("d", "123");
    
        Map<String, Object> m4 = new HashMap<String, Object>();
        m4.put("b", "123");
        m4.put("a", "abc");
        m4.put("c", "123");
        m4.put("d", "123");
    
        List<Map<String, Object>> a = new ArrayList<>();
        a.add(m1);
        a.add(m3);
    
        List<Map<String, Object>> b = new ArrayList<>();
        b.add(m2);
        b.add(m4);
    
        boolean c = isEqualCollection(a, b);
        System.out.println(c);
    
    }

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原创不易，转载自：[https://blog.csdn.net/qq\_28245087/article/details/93187235][https_blog.csdn.net_qq_28245087_article_details_93187235]

[Java_2021]: https://blog.csdn.net/guorui_java/article/details/112391105
[Java]: https://blog.csdn.net/guorui_java/article/details/107599455
[springBoot]: https://blog.csdn.net/guorui_java/article/details/104137262
[Link 1]: https://blog.csdn.net/guorui_java/article/details/106217472
[Java_23]: https://blog.csdn.net/guorui_java/article/details/104026988
[Java 1]: https://blog.csdn.net/guorui_java/article/details/107803283
[https_blog.csdn.net_qq_28245087_article_details_93187235]: https://blog.csdn.net/qq_28245087/article/details/93187235