请用Java实现，现有一个对象A，存有属性a,b,c,d,e,f六个属性，根据a,b,c三个属性可以确定唯一值，现List<A>中，d,e的值可能为空，N或者Y，现需要将相同的a,b,c对象值合并，并且d,e存在N则赋值为N，否则保持原样，返回List<A>

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

public class MergeA {
    private String a;
    private String b;
    private String c;
    private String d;
    private String e;
    private String f;

    // Constructor
    public MergeA(String a, String b, String c, String d, String e, String f) {
        this.a = a;
        this.b = b;
        this.c = c;
        this.d = d;
        this.e = e;
        this.f = f;
    }

    // Getter and Setter methods
    public String getA() {
        return a;
    }

    public void setA(String a) {
        this.a = a;
    }

    public String getB() {
        return b;
    }

    public void setB(String b) {
        this.b = b;
    }

    public String getC() {
        return c;
    }

    public void setC(String c) {
        this.c = c;
    }

    public String getD() {
        return d;
    }

    public void setD(String d) {
        this.d = d;
    }

    public String getE() {
        return e;
    }

    public void setE(String e) {
        this.e = e;
    }

    public String getF() {
        return f;
    }

    public void setF(String f) {
        this.f = f;
    }

    // Method to merge list of objects based on a,b,c values
    public static List<MergeA> merge(List<MergeA> list) {
        List<MergeA> result = new ArrayList<>();
        Map<String, MergeA> map = new HashMap<>();
        for (MergeA a : list) {
            String key = a.getA() + a.getB() + a.getC();
            if (map.containsKey(key)) {
                MergeA existing = map.get(key);
                if (a.getD().equals("N")) {
                    existing.setD("N");
                }
                if (a.getE().equals("N")) {
                    existing.setE("N");
                }
            } else {
                map.put(key, a);
                result.add(a);
            }
        }
        return result;
    }

    // Main method to test
    public static void main(String[] args) {
        MergeA a1 = new MergeA("A", "B", "C", "Y", "N", "F");
        MergeA a2 = new MergeA("A", "B", "C", null, "Y", "F");
        MergeA a3 = new MergeA("D", "E", "F", "N", "N", "F");
        MergeA a4 = new MergeA("D", "E", "F", "N", "Y", "F");

        List<MergeA> list = new ArrayList<>();
        list.add(a1);
        list.add(a2);
        list.add(a3);
        list.add(a4);

        List<MergeA> mergedList = MergeA.merge(list);

        for (MergeA a : mergedList) {
            System.out.println(a.getA() + ", " + a.getB() + ", " + a.getC() + ", " + a.getD() + ", " + a.getE() + ", " + a.getF());
        }
    }
}

代码主要实现了一个 MergeA 类，其中包含了 a,b,c,d,e,f 六个属性，并提供了一个 merge 方法，用于将 List<MergeA> 中相同的 a,b,c 对象值合并，并且将 d,e 存在 N 则赋值为 N，否则保持原样，最后返回合并后的 List<MergeA> 列表。

在 main 方法中，我们创建了四个 MergeA 对象，并将它们添加到列表 list 中。然后，我们调用 MergeA.merge(list) 方法，将会得到合并后的列表 mergedList。最后，我们循环遍历 mergedList，打印出每个 MergeA 对象的属性值。