将重叠的时间范围与其权重合并答案

【问题标题】：Merging overlapping time ranges with their weights将重叠的时间范围与其权重合并
【发布时间】：2019-10-23 06:51:32
【问题描述】：

我有一个 DateRange 模型，如下所示：

public class DateRange
{
    public DateTime start { get; set; }
    public DateTime end { get; set; }
    public int weight { get; set; }
}

这里是列表：

[
  {
    "start": 1, "end": 3, "weight": 1
  },
  {
    "start": 10, "end": 20, "weight": 3
  },
  {
    "start": 12, "end": 15, "weight": 2
  },
  {
    "start": 22, "end": 26, "weight": 7
  },
  {
    "start": 25, "end": 30, "weight": 4
  }
]

我想合并列表。合并列表的重叠时间范围的权重必须相加。

结果样本：

[
  {
    "start": 1, "end": 3, "weight": 1
  },
  {
    "start": 10, "end": 12, "weight": 3
  },
  {
    "start": 12, "end": 15, "weight": 5
  },
  {
    "start": 15, "end": 20, "weight": 3
  },
  {
    "start": 22, "end": 25, "weight": 7
  },
  {
    "start": 25, "end": 26, "weight": 11
  },
  {
    "start": 26, "end": 30, "weight": 4
  }
]

【问题讨论】：

你试过什么？你面临什么问题？
这是作业吗？这看起来是一项相当乏味的任务。

标签： c# .net algorithm

【解决方案1】：

public class DataComparer : IComparer<Data>
{
    public int Compare(Data d1, Data d2)
    {
        if (d1.start == d2.start)
        {
            return d1.end.CompareTo(d2.end);
        }
        return d1.start.CompareTo(d2.start);
    }
}

public List<Data> resolveConflicts(List<Data> list)
{
    list.Sort(new DataComparer());
    for (int i = 0; i < list.Count - 1; i++)
    {
        var current = list[i];
        var next = list[i + 1];
        if (i > 1)
        {
            var pre = list[i - 1];
            if (pre.end == current.start && pre.weight == current.weight) // 10 - 12 > 12 - 14
            {
                i = i - 2;
                continue;
            }
        }
        if (next.start < current.end)
        {
            if (next.end < current.end) //10 - 20 > 12 - 15
            {
                list.Add(new Data(next.end, current.end, current.weight));
                list.Sort(new DataComparer());

                current.end = next.start;
                next.weight += current.weight;
                i--;
            }
            else if (next.end == current.end) // 1 - 5 // 2 - 5
            {
                next.weight += current.weight;
                if (current.start == next.start)
                {
                    list.Remove(current);
                    i--;
                }
                else
                {
                    current.end = next.start;
                }
            }
            else if (next.end > current.end) //1 - 5 > 4 - 6
            {
                if (next.start == current.start) // 1 - 6 > 1 - 7
                {
                    current.weight += next.weight;
                    next.start = current.end;

                    list.Sort(new DataComparer());
                    i--;
                }
                else
                {
                    list.Add(new Data(next.start, current.end, next.weight + current.weight));

                    var temp = current.end;
                    current.end = next.start;
                    next.start = temp;

                    list.Sort(new DataComparer());
                    i--;
                    continue;
                }
            }
        }
        else if (next.start == current.end && next.weight == current.weight) // 1 - 5 > 5 - 7
        {
            current.end = next.end;
            list.Remove(next);
            i--;
        }

    }

    return list;
}

【讨论】：