本文整理汇总了C#中Queue.Reverse方法的典型用法代码示例。如果您正苦于以下问题：C# Queue.Reverse方法的具体用法？C# Queue.Reverse怎么用？C# Queue.Reverse使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Queue的用法示例。
在下文中一共展示了Queue.Reverse方法的15个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: GetShortestSequence
private static IEnumerable<int> GetShortestSequence(int start, int end)
{
var sequence = new Queue<int>();
while (start <= end)
{
sequence.Enqueue(end);
if (end / 2 > start)
{
if (end % 2 == 0)
{
end /= 2;
}
else
{
end--;
}
}
else
{
if (end - 2 >= start)
{
end -= 2;
}
else
{
end--;
}
}
}
return sequence.Reverse();
}示例2: DisplayShortestSequence
private static void DisplayShortestSequence(int start, int target)
{
var sequence = new Queue<int>();
while (start <= target)
{
sequence.Enqueue(target);
if (target / 2 >= start)
{
if (target % 2 == 0)
{
target /= 2;
}
else
{
target--;
}
}
else
{
if (target - 2 >= start)
{
target -= 2;
}
else
{
target--;
}
}
}
Console.WriteLine(string.Join(" -> ", sequence.Reverse()));
}示例3: Main
public static void Main()
{
int start = 5;
int end = 16;
Queue<int> quene = new Queue<int>();
while (end >= start)
{
quene.Enqueue(end);
if (end / 2 >= start)
{
if (end % 2 == 0)
{
end /= 2;
}
else
{
end--;
}
}
else if (end - 2 >= start)
{
end -= 2;
}
else
{
end--;
}
}
Console.WriteLine(string.Join(", ", quene.Reverse()));
}示例4: FindShortestSequenceOfOperations
public static void FindShortestSequenceOfOperations(int start, int end)
{
Queue<int> steps = new Queue<int>();
int current = end;
while (current >= start)
{
steps.Enqueue(current);
if (current / 2 >= start)
{
if (current % 2 == 0)
{
current /= 2;
}
else
{
current -= 1;
}
}
else if (current - 2 >= start)
{
current -= 2;
}
else
{
current -= 1;
}
}
Console.WriteLine(string.Join(" -> ", steps.Reverse()));
}示例5: reconstruct_path
private Queue<Vector2> reconstruct_path(Dictionary<Node, Node> came_from, Node current)
{
Queue<Vector2> path = new Queue<Vector2>();
path.Enqueue(current.place);
while(came_from.ContainsKey(current))
{
current = came_from[current];
path.Enqueue(current.place);
}
return new Queue<Vector2>( path.Reverse());
}示例6: Main
public static void Main()
{
Console.Write("Enter sequence start: ");
int startNumber = int.Parse(Console.ReadLine());
Console.Write("Enter sequence end: ");
int endNumber = int.Parse(Console.ReadLine());
Queue<int> operations = new Queue<int>();
while (startNumber <= endNumber)
{
operations.Enqueue(endNumber);
if (endNumber / 2 >= startNumber)
{
if (endNumber % 2 == 0)
{
endNumber /= 2;
}
else
{
endNumber--;
}
}
else
{
if (endNumber - 2 >= startNumber)
{
endNumber -= 2;
}
else
{
endNumber--;
}
}
}
Console.WriteLine(string.Join(" -> ", operations.Reverse()));
}示例7: Main
static void Main()
{
var sequence = new Queue<int>();
Console.WriteLine("Enter N: ");
var start = int.Parse(Console.ReadLine());
Console.WriteLine("Enter M: ");
var end = int.Parse(Console.ReadLine());
var numbers = new Queue<int>();
while (start <= end)
{
numbers.Enqueue(end);
if (end / 2 >= start)
{
if (end % 2 == 0)
{
end /= 2;
}
else
{
end--;
}
}
else
{
if (end - 2 >= start)
{
end -= 2;
}
else
{
end--;
}
}
}
Console.WriteLine(string.Join(" -> ", numbers.Reverse()));
}示例8: Main
public static void Main()
{
//int n = ConsoleTool.ReadInteger();
//int m = ConsoleTool.ReadInteger();
int n = 5;
int m = 16;
Queue<int> queue = new Queue<int>();
while (n <= m)
{
queue.Enqueue(m);
if (m / 2 >= n)
{
if (m % 2 == 0)
{
m /= 2;
}
else
{
m--;
}
}
else
{
if (m - 2 >= n)
{
m -= 2;
}
else
{
m--;
}
}
}
Console.WriteLine(string.Join(",", queue.Reverse()));
}示例9: Main
public static void Main()
{
Console.Write("Enter n: ");
int n = int.Parse(Console.ReadLine());
Console.Write("Enter m: ");
int m = int.Parse(Console.ReadLine());
Queue<int> queue = new Queue<int>();
while (n <= m)
{
queue.Enqueue(m);
if (m / 2 >= n)
{
if (m % 2 == 0)
{
m /= 2;
}
else
{
m--;
}
}
else
{
if (m - 2 >= n)
{
m -= 2;
}
else
{
m--;
}
}
}
Console.WriteLine(string.Join(" --> ", queue.Reverse()));
}示例10: Main
public static void Main()
{
var result = new Queue<int>();
int start = 5;
int end = 16;
while (start <= end)
{
result.Enqueue(end);
if (end / 2 >= start)
{
if (end % 2 == 0)
{
end /= 2;
}
else
{
end--;
}
}
else
{
if (end - 2 >= start)
{
end -= 2;
}
else
{
end--;
}
}
}
Console.WriteLine(string.Join(" -> ", result.Reverse()));
}示例11: StartTempLogging
//.........这里部分代码省略.........
rs = BuildDatabase(alldata.Where(x => x.Log).ToList());
}
initialized = true;
}
if (logmode)
{
RecordSample(rs, alldata);
}
if (dumpmode)
{
using (var w = new StringWriter())
{
mon.DumpRawByteData(w);
File.AppendAllText("W83627DHG.log", w.ToString());
}
using (var w = new StringWriter())
{
mon.DumpSIOByteData(w);
File.AppendAllText("Super-IO.log", w.ToString());
}
//using (var w = new StringWriter())
//{
// asp0800.DumpRawByteData(w);
// File.AppendAllText("ASP0800b.log", w.ToString());
//}
using (var w = new StringWriter())
{
asp0800.DumpRawWordData(w);
File.AppendAllText("ASP0800w.log", w.ToString());
}
}
if (consolemode)
{
Console.Title = string.Format("TempServe - Running for: {3}d {0:00}:{1:00}:{2:00}",
elapsed.Hours, elapsed.Minutes, elapsed.Seconds, elapsed.Days);
while (avg5.Count >= 60)
{
avg5.Dequeue();
}
var hashdata = alldata.ToDictionary(x => x.Name, x => x); ;
avg5.Enqueue(hashdata);
if (min == null)
{
min = hashdata;
}
else
{
min = HashMap((x, y) => x.Value > y.Value ? y : x, min, hashdata);
}
if (max == null)
{
max = hashdata;
}
else
{
max = HashMap((x, y) => x.Value < y.Value ? y : x, max, hashdata);
}
using (var output = new StringWriter())
{
output.WriteLine("{0,-18}{1,10}{2,10}{3,10}{4,10}{5,10}{6,10}",
"Sensor", "Current", "Min", "Max", "Avg(10)", "Avg(30)", "Avg(60)");
foreach (var s in alldata)
{
var avg = avg5.Reverse().ToArray();
var fs = BuildFormatString(s);
output.WriteLine(fs, s.Name, s.Value,
min[s.Name].Value,
max[s.Name].Value,
avg.Take(10).Select(x => x[s.Name]).Average(x => x.Value),
avg.Take(30).Select(x => x[s.Name]).Average(x => x.Value),
avg.Take(60).Select(x => x[s.Name]).Average(x => x.Value)
);
}
Console.Clear();
Console.Write(output.ToString());
}
}
var taken = Math.Min((sw.Elapsed - elapsed).TotalMilliseconds, 100) + 3;
Thread.Sleep(1000 - (int)taken);
}
}
}示例12: createPathFromQueue
/// <summary>
/// Convert the path from dijkstra algorithm to fit natural version from start node to finish node
/// </summary>
private static void createPathFromQueue(List<List<int>> visited, out List<int> path, int node, int startNode)
{
Queue<int> q = new Queue<int>();
Queue<int> qtemp = new Queue<int>();
q.Enqueue(node);
while (true)
{
node = visited[node][0];
q.Enqueue(node);
if (node == startNode)
{
break;
}
}
path = q.Reverse().ToList<int>();
}示例13: FindShortestSeqOfOperationFromNToM
// 10.
public void FindShortestSeqOfOperationFromNToM(int start, int end)
{
var queue = new Queue<int>();
while (start != end)
{
queue.Enqueue(end);
if (end / 2 >= start)
{
if (end % 2 == 0)
{
end /= 2;
}
else
{
end--;
}
}
else
{
if (end - 2 >= start)
{
end -= 2;
}
else
{
end--;
}
}
}
queue.Enqueue(start);
Console.WriteLine(string.Join(", ", queue.Reverse()));
}示例14: reconstruct_path
void reconstruct_path(
Dictionary<Path_Node<Tile>, Path_Node<Tile>> Came_From,
Path_Node<Tile> current
)
{
// So at this point, current IS the goal.
// So what we want to do is walk backwards through the Came_From
// map, until we reach the "end" of that map...which will be
// our starting node!
Queue<Tile> total_path = new Queue<Tile>();
total_path.Enqueue(current.data); // This "final" step is the path is the goal!
while( Came_From.ContainsKey(current) ) {
// Came_From is a map, where the
// key => value relation is real saying
// some_node => we_got_there_from_this_node
current = Came_From[current];
total_path.Enqueue(current.data);
}
// At this point, total_path is a queue that is running
// backwards from the END tile to the START tile, so let's reverse it.
path = new Queue<Tile>( total_path.Reverse() );
}示例15: ComparePad
private bool ComparePad(Queue<cmd> queue)
{
int ipos = padtag.Count - 1;
if (ipos < 0) return false;
int iseek = 1;
int lastpad = 0;
int pd = 0;
foreach (cmd p in queue.Reverse())
{
if (p.padtag == lastpad) continue; //1.输入冗余过滤
lastpad = p.padtag;
if (p.padtag == 5) //2.无输入冗余过滤
{
if (padtag[ipos] == 5) ipos--;//输入要求空
if (ipos < 0) return true;
continue;//无方向，跳过
}
if (ipos < padtag.Count - 1)
{
if (padtag[ipos + 1] == p.padtag) //3.简化操作判定
{//相容引起的重复可以跳
if(pd>0)
continue;//重复方向，跳过
}
}
pd = PadDist(padtag[ipos], p.padtag);
if (pd > 1) return false;
ipos--;
if (ipos < 0) return true;
if (pd == 0)//相等
{
iseek = 0;
//ipos--;
}
else//相容 //3.简化操作判定，斜方向可以简化两个方向
{
int irpd=0;
if(ipos+2<padtag.Count)
irpd = padtag[ipos + 2]; //3.1同一个四分之一圆里来回不能简化
if (iseek == 0)
{
pd = PadDist(padtag[ipos], p.padtag);
if (pd <= 1)
{
iseek = 0;
if (padtag[ipos] != irpd)
{
ipos--;
if (ipos < 0) return true;
}
}
}
iseek++;
if (iseek > 1) iseek = 0;
}
}
return false;
}