public DataResult <NumRange> GetNumRange(string NumRangeCode)
{
try
{
using (var db = new SocietyDBEntities())
{
NumRange objNumRange = db.NumRanges.Where(x => x.numRangeCode == NumRangeCode && x.isDeleted == false).FirstOrDefault();
if (objNumRange != null)
{
var NewNumber = objNumRange.currentNumber + 1;
if (NewNumber > objNumRange.startNumber && NewNumber < objNumRange.endNumber)
{
objNumRange.currentNumber = NewNumber;
db.Entry(objNumRange).State = System.Data.EntityState.Modified;
db.SaveChanges();
var NewNumberString = NewNumber.ToString().PadLeft(8, '0');
NewNumberString = NumRangeCode + NewNumberString;
Result.StringResult = NewNumberString;
Result.Message = "NumRange Generated Successfully.";
Result.MsgType = "S";
}
else
{
Result.Message = "Numrange Exceeds.Kindly Contact Administration.";
Result.MsgType = "E";
}
}
else
{
Result.Message = "Numrange Not Found.Kindly Contact Administration.";
Result.MsgType = "E";
}
}
}
catch (Exception ex)
{
Result.Message = ex.Message;
Result.MsgType = "E";
}
return(Result);
}
public int[][] Merge(int[][] intervals)
{
/*
* 将多个片段做个处理，即，当片段重叠，那么就直接合并成一个片段
* 思路：
* 1.将片段按照区间开始的大小来排序
* 2.然后外层循环依次遍历所有的片段，内层循环查找并合并重叠片段
*
* 时间复杂度：排序O(nlogn)+遍历O(n)，所以最后是，O(nlogn)
* 空间复杂度：没有利用额外的空间，所以是，O(1)
*
* 考察点：
* 1.排序、数组
*/
//1.对区间片段排序
List <NumRange> ranges = new List <NumRange>();
for (int i = 0; i < intervals.GetLength(0); i++)
{
ranges.Add(new NumRange()
{
Min = intervals[i][0],
Max = intervals[i][1]
});
}
if (!ranges.Any())
{
return(intervals);
}
var orderedRanges = ranges.OrderBy(i => i.Min).ToList();
//2.利用内外循环的方式做遍历
var mergedRanges = new List <NumRange>();
for (int j = 0; j < orderedRanges.Count; j++)
{
NumRange newRange = new NumRange();
newRange.Min = orderedRanges[j].Min;
newRange.Max = orderedRanges[j].Max;
while (j + 1 < orderedRanges.Count && orderedRanges[j + 1].Min <= newRange.Max)
{
if (orderedRanges[j + 1].Max > newRange.Max)
{
newRange.Max = orderedRanges[j + 1].Max;
}
j++;
}
mergedRanges.Add(newRange);
}
//3.返回需要的结果
int[][] forReturn = new int[mergedRanges.Count][];
for (int k = 0; k < mergedRanges.Count; k++)
{
forReturn[k] = new int[] { mergedRanges[k].Min, mergedRanges[k].Max }
}
;
return(forReturn);
}
}
