我有做功课：写一个程序，找到数组中相等元素的最大序列。示例：{2,1,1,2,3,3,2,2,2,1} = {2,2,2}。 我想出了这个：
Console.WriteLine("Enter array lenght");
int arrLenght = int.Parse(Console.ReadLine());
int[] arr = new int[arrLenght];
Console.WriteLine("Enter array elements");
for (int i = 0; i < arr.Length; i++)
{
arr[i] = int.Parse(Console.ReadLine());
}
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == arr[i + 1] && arr[i] == arr[i + 2])
{
Console.WriteLine("Maximal sequence of numbers is: {0},{1},{2}",arr[i],arr[i+1],arr[i+2]);
break;
}
}
仅当序列正好是3个数字时才有效。我必须搜索数组并找到最大的序列，但我不知道如何编码。我很抱歉，如果问题很愚蠢，但我是新手，我无法在其他任何地方找到解决方案。感谢
答案 0 :(得分：3)
如果您正在寻找优雅，请使用Linq
var seq = new int[] {2, 1, 1, 2, 3, 3, 2, 2, 2, 1};
int[] max = seq.Select((n, i) => new { Value = n, Index = i})
.OrderBy(s => s.Value)
.Select((o, i) => new { Value = o.Value, Diff = i - o.Index } )
.GroupBy(s => new { s.Value, s.Diff})
.OrderByDescending(g => g.Count())
.First()
.Select(f => f.Value)
.ToArray();
这就是我♥Linq
的原因答案 1 :(得分：2)
Linq：
int count = seq.Count();
int[] maxSeq = seq
.Select((i, index) => new{
Item = i, index,
PrevEqual = index == 0 || seq.ElementAt(index - 1) == i,
NextEqual = index == count - 1 || seq.ElementAt(index + 1) == i,
})
.Where(x => x.PrevEqual || x.NextEqual)
.GroupBy(x => x.Item)
.OrderByDescending(g => g.Count())
.First().Select(x => x.Item).ToArray();
<强>解释
bool属性的匿名类型，该属性指示它是否与上一个Where GroupBy具有相同值的元素答案 2 :(得分：0)
如果没有完全给出你问题的答案，我可以建议你看看LINQ，因为肯定会有一些简单的方法来解决这个问题。您可以先查看可以为您提供答案的Group，Count，OrderBy和Select扩展程序。
答案 3 :(得分：0)
由于这是一项功课，可能需要建立算法
int[] arr = new int[30];//your array
Random rand = new Random(100);
int maxCount = 0, curCount, value = 0;
for (int i = 0; i < arr.Length; i++)
arr[i] = rand.Next(15);//fill the aray with random values
arr = arr.OrderBy(a => a).ToArray();
for (int i = 0; i < arr.Length; i++)
{
curCount = 1;//found new value. and now count == 1
for (int j = i+1/*search from next array element*/;
j < arr.Length-1/*to the end*/;
j++)
{
if (arr[i] == arr[j])
curCount++;//counts the count
else
break;//met new value
}
if (curCount > maxCount)//we've found new sequence
{
maxCount = curCount;//new sequence length
value = arr[i];//sequence values
i += maxCount;//we don't need to watch the sequence again
}
}
我现在手上还没有VS来检查这个，所以我希望有效吗=）无论如何有一个想法
答案 4 :(得分：0)
尝试使用此方法：
int MaximimalSequence<T>(IList<T> list, out T value)
{
T aux = default(T);
value = default(T);
int max = 0, hist = 0;
bool first = true;
foreach (var i in list)
{
if (!first && aux.Equals(i))
{
max++;
}
else
{
first = false;
max = 1;
}
if (hist < max)
{
hist = max;
value = i;
}
aux = i;
}
return hist;
}
要打电话：
int value;
var maximumSequence = MaximimalSequence<int>(new List<int> { 2, 1, 1, 2, 3, 3, 2, 2, 2, 1 }, out i);
答案 5 :(得分：0)
这是一个可以通过序列的单次迭代来解决的问题。重要的是要确保算法适用于所有情况，包括在最大序列位于序列末尾时。
private static IEnumerable<int> GetMaxSequence(IList<int> seq)
{
if (seq == null || seq.Count == 0)
{
return new List<int>();
}
int value = seq[0];
int currentSequenceStartIndex = 0;
int currentSequenceLength = 1;
int maxSequenceStartIndex = 0;
int maxSequenceLength = 0;
for (int i = 1; i < seq.Count; i++)
{
if (seq[i] == value)
{
currentSequenceLength++;
continue;
}
if (currentSequenceLength > maxSequenceLength)
{
maxSequenceLength = currentSequenceLength;
maxSequenceStartIndex = currentSequenceStartIndex;
}
currentSequenceStartIndex = i;
currentSequenceLength = 1;
value = seq[i];
}
if (currentSequenceLength > maxSequenceLength)
{
maxSequenceLength = currentSequenceLength;
maxSequenceStartIndex = currentSequenceStartIndex;
}
return seq.Skip(maxSequenceStartIndex).Take(maxSequenceLength);
}
答案 6 :(得分：0)
// create two list holding ints. One for the temporary value and one for the longest sequence
List<int> longestSequence = new List<int>();
List<int> temp = new List<int>();
// create count to count how many elements are holding the same value
// and counter to assign this value when max is reached
int count = 0;
int counter = 0;
// with for loop compare every element with the elements that follow it
for (int i = 0; i < arr.Length - 1; i++)
{
int nextElement = i+1; // element that follows
count = 0; // ignore for now see the end of the for loop
temp.Clear(); // ignore for now see the end of the for loop
temp.Add(arr[i]); // add the compared element in to the temp list
// while the value is the same as following element add this to the temporary list
// and add count (count++)
while (arr[i] == arr[nextElement])
{
temp.Add(arr[nextElement]);
nextElement++;
count++;
}
// after that see if the count is bigger than counter (maximum amount of elements so far)
// it is set to 0 at the beginning
if (count > counter)
{
longestSequence.Clear();
// if it is bigger assign count value to counter
counter = count;
// and copy the temporary list to the longestSequence list
for (int k = 0; k < temp.Count; k++)
{
longestSequence.Add(temp[k]);
}
}
// at the beggining of the for loop the count will be set to 0 again
// and the temporary list will be cleared
}
Console.WriteLine();
// print the longestSequence list
foreach (int element in longestSequence)
{
Console.Write(element + " ");
}
Console.WriteLine();
答案 7 :(得分：0)
SELECT
locations."name",
contacts.name,
array_agg(roles."name") AS "Roles"
FROM locations INNER JOIN contacts ON locations."id" = contacts.location_id
INNER JOIN "public".memberships ON "public".memberships.contact_id = contacts."id"
INNER JOIN roles ON "public".memberships.role_id = roles."id"
group by locations."name", contacts.name
答案 8 :(得分：0)
//程序员：Hamza Jany // P - Lang = C＃
static int num;
static void Main(string[] args)
{
int[] array;
Console.Write("enter the size of array : ");
int input = int.Parse(Console.ReadLine());
int[] temp;
temp = new int[10];
int a = 0;
int count = 0;
array = new int[input];
for (int i = 0; i < input; i++)
{
Console.Write("enter value ( " + i + ", " + input+") :");
array[i] = int.Parse(Console.ReadLine());
}
for (int i = 0; i < int.MaxValue; i++)
{
for (int j = 0; j < array.Length; j++)
{
if (i == array[j])//compare i with array
{
count++;
if (a < count)//if greator found
{
a = count;
num = i;
}
}
else
{
count = 0;
}
}
}
Console.WriteLine(num +" repeated " + a +" times");
Console.ReadKey();
}
答案 9 :(得分：0)
static void Main(string[] args)
{
int[] array1 = { 1, 1, 1, 2, 3, 2, 2, 2, 2, 1 };
int start = 0;
int length = 0;
int bestStart = 0;
int bestLength = int.MinValue;
for (int i = 0; i < array1.Length - 1; i++)
{
if ((i == 0) || (array1[i] != array1[i - 1]))
{
start = i;
}
if(array1[start] == array1[start + 1])
{
length++;
if (length > bestLength)
{
bestLength = length;
bestStart = start;
}
}
else
{
length = 0;
}
}
Console.Write("The best sequence is :");
for (int i = bestStart; i < bestLength + bestStart; i++)
{
Console.Write(" " + array1[i]);
}
}
/*
This will give : The best sequence is : 2 2 2 2
*/
答案 10 :(得分：0)
以下是我对C＃中问题的解决方案。我尝试尽可能减少代码，使其更具可读性和可理解性。
public static void Main(string[] args)
{
int[] intList = new int[] { 1, 1, 2, 2, 3, 3, 3, 4, 4, 5 };
foreach (var item in intList) {
Console.Write(item + " ");
}
#region Calculating sequence and printing details
int length = 1, start = 0, finalIndx = 0,
bigSeqNum = 0, finalLen = 0;
for (int i = 1; i < intList.Length; i++) {
if (intList[i] == intList[start]) {
length++;
if (length > finalLen) {
finalLen = length;
finalIndx = start;
bigSeqNum = intList[start];
}
} else {
start = i;
length = 1;
}
}
Console.WriteLine("\nBig Seq. Num: " + bigSeqNum);
Console.WriteLine("Start index: " + finalIndx);
Console.WriteLine("Length: " + finalLen);
Console.WriteLine();
#endregion
}
答案 11 :(得分：-1)
static void Main(string[] args)
{
Console.WriteLine("enter length of the array");
int n = int.Parse(Console.ReadLine());
int[] myarray = new int[n];
for (int i = 0; i < n ; i++)
{
Console.WriteLine("enter value of array" + " " + i);
myarray[i]=int.Parse(Console.ReadLine());
}
int length = 1;
int start = 0;
int bestlength=0;
int beststart=0;
for (int i = 1; i < n; i++)
{
if (myarray[i - 1] == myarray[i])
{
length++;
continue;
}
if (bestlength<length)
{
bestlength = length;
beststart = start;
}
length = 1;
start = i;
}
Console.WriteLine("the best sequence is");
for (int j = beststart; j < bestlength + beststart; j++)
{
Console.Write(myarray[j] + ",");
}
}
}
}