如何将对象列表项插入指定分组末尾并保留原有排序

有两个Author对象列表authors1和authors2，其中authors1已按SDate字段排序。需求是将authors2中的项添加到authors1中，保持原有排序的同时，所有新增项需放在对应Book分组的末尾。

现有代码 #

IList<Author> authors1 = new List<Author>
{
    new Author { Book = "c#", Code="A11" , SDate = 1 },
    new Author { Book = "c#", Code="A22" , SDate = 1 },
    new Author { Book = "c#", Code="A31" , SDate = 1 },

    new Author { Book = "js", Code="B43" , SDate = 2 },
    new Author { Book = "js", Code="B33" , SDate = 2 },
    new Author { Book = "js", Code="B41" , SDate = 2 },

    new Author { Book = "java", Code="C27", SDate = 3 },
    new Author { Book = "java", Code="C33", SDate = 3 },
    new Author { Book = "java", Code="C78", SDate = 3 }
};

IList<Author> authors2 = new List<Author>
{
    new Author { Book = "c#", Code = "A21" },
    new Author { Book = "java", Code = "C23" }
};

预期结果 #

IList<Author> authors1 = new List<Author>
{
    new Author { Book = "c#", Code="A11", SDate = 1 },
    new Author { Book = "c#", Code="A22", SDate = 1 },
    new Author { Book = "c#", Code="A31", SDate = 1 },
    new Author { Book = "c#", Code="A21", SDate = 1 },  // 新增项添加到该分组末尾

    new Author { Book = "js", Code="B43", SDate = 2 },
    new Author { Book = "js", Code="B33", SDate = 2 },
    new Author { Book = "js", Code="B41", SDate = 2 },

    new Author { Book="java", Code="C27", SDate=3 },
    new Author { Book="java", Code="C33", SDate=3 },
    new Author { Book="java", Code="C78", SDate=3 },
    new Author { Book="java", Code="C23", SDate=3 }  // 新增项添加到该分组末尾
};

当前尝试的代码片段 #

IList<Author> response = new List<Author>();

foreach (var author in authors2.GroupBy(x => x.Book).ToList())
{
    foreach (var item in author)
    {

解决方案 #

方法一：分组拼接法 #

先将authors1按Book分组，把authors2的项补充到对应分组（同步设置SDate），最后按原SDate顺序拼接所有分组，高效实现需求：

// 把authors1按Book分组，保存每个分组的SDate和作者列表
var groupedAuthors1 = authors1.GroupBy(a => a.Book)
                              .ToDictionary(g => g.Key, g => new 
                              { 
                                  SDate = g.First().SDate, 
                                  Authors = g.ToList() 
                              });

// 处理authors2中的每个项，补充到对应分组
foreach (var author in authors2)
{
    if (groupedAuthors1.TryGetValue(author.Book, out var groupInfo))
    {
        // 设置和同分组一致的SDate
        author.SDate = groupInfo.SDate;
        groupInfo.Authors.Add(author);
    }
    else
    {
        // 处理authors1中没有的Book分组（可选逻辑，这里设置SDate为现有最大值+1）
        var maxSDate = groupedAuthors1.Values.Max(v => v.SDate);
        author.SDate = maxSDate + 1;
        groupedAuthors1.Add(author.Book, new 
        { 
            SDate = author.SDate, 
            Authors = new List<Author> { author } 
        });
    }
}

// 按原SDate顺序排序分组，拼接成最终列表
var result = groupedAuthors1.Values.OrderBy(v => v.SDate)
                                   .SelectMany(v => v.Authors)
                                   .ToList();

方法二：遍历插入法 #

遍历authors1的同时跟踪当前Book分组，在分组切换时插入对应authors2的项，严格保持原分组内的顺序：

IList<Author> response = new List<Author>();
// 把authors2按Book分组，方便快速查找
var authors2Grouped = authors2.GroupBy(a => a.Book)
                              .ToDictionary(g => g.Key, g => g.ToList());

string currentBook = null;

foreach (var author in authors1)
{
    if (currentBook != author.Book)
    {
        // 切换分组时，先把上一个分组的新增项加入列表（第一次循环跳过）
        if (currentBook != null && authors2Grouped.TryGetValue(currentBook, out var itemsToAdd))
        {
            var sDate = authors1.First(a => a.Book == currentBook).SDate;
            foreach (var item in itemsToAdd)
            {
                item.SDate = sDate;
                response.Add(item);
            }
        }
        currentBook = author.Book;
    }
    response.Add(author);
}

// 处理最后一个分组的新增项
if (currentBook != null && authors2Grouped.TryGetValue(currentBook, out var lastItemsToAdd))
{
    var sDate = authors1.First(a => a.Book == currentBook).SDate;
    foreach (var item in lastItemsToAdd)
    {
        item.SDate = sDate;
        response.Add(item);
    }
}

两种方法都能满足需求：保持authors1原有排序，新增项放在对应Book分组的末尾，同时自动同步SDate字段。

内容的提问来源于stack exchange，提问作者suc pepe