Case: The procedure of sorting odd even-numbered is implemented by generics. Int[] nums={1,2,3,4,5,6,7,8,9}; Odd on the left even on the right
list<int> list = new List<int> () {1, 2, 3, 4, 5, 6, 7, 8, 9};
list<int> Odd = new list<int> ();//Odd
list<int> even = new list<int> ();//Even
for (int i = 0; i < list. Count; i++)
//{
If (List[i]% 2! = 0)
// {
Even.add (List[i]);
// }
Else
// {
Odd.add (List[i]);
// }
//}
List. AddRange (ODD);
List. AddRange (even);
StringBuilder sb = new StringBuilder (); Variable string
for (int i = 0; i < list. Count; i++)
//{
Sb. Append (List[i]);
//}
Console.WriteLine (SB);
Console.readkey ();
Exercise 1: Return the odd number in an int array to a new int array.
The odd number in the array is taken out into a collection, and the collection is eventually converted to a group.
Int[] Nums = {1, 2, 3, 4, 5, 6, 7, 8, 9};
list<int> listodd = new list<int> ();
for (int i = 0; i < Nums. Length; i++)
//{
If (Nums[i]% 2! = 0)
// {
Listodd.add (Nums[i]);
// }
//}
Then convert the collection to the appropriate array
int[] Newnums = Listodd.toarray ();
for (int i = 0; i < newnums.length; i++)
//{
Console.WriteLine (Newnums[i]);
//}
Console.readkey ();
Exercise 2: Remove the maximum number from the list<int> of an integer (find the maximum).
list<int> list = new List<int> () {1, 2, 3, 4, 5, 6, 3, 8, 7, 9};
int temp=list[0];
for (int i = 0; i < list. Count; i++)
//{
if (list[i]>temp)
// {
Temp=list[i];
// }
//}
Console.WriteLine ("The maximum value in this array is {0}", temp);
Console.readkey ();
Exercise: Convert 123 to: one and three. Dictionary<char,char>
"1.,122,334,455,667,79e,+17"
Console.WriteLine ("Please enter a number");
String number = Console.ReadLine ();
String str = "1 One 2 II 3 three 4 restaurant 5 Wu Lu 6 seven 7 BA 8 JIU 9";
Dictionary<int, char> dic = new Dictionary<int, char> ();
string[] parts = str. Split (new char[] {", StringSplitOptions.RemoveEmptyEntries.ToString ()});
for (int i = 0; i < parts. Length; i++)
////{
Dic. ADD (Parts[i][0], parts[i][1]);
////}
StringBuilder sb = new StringBuilder ();
for (int i = 0; i < number. Length; i++)
////{
Sb. Append (Dic[number[i]);
////}
Console.WriteLine (sb.) ToString ());
Console.readkey ();
Exercise: Calculate the number of occurrences of each character in a string (interview questions). "Welcome to Chinaworld", Case insensitive, print "W2" "E 2" "O 3" ...
Tip: There are many static methods for Dictionary<char,int>,char. Char. Isletter ()
String str = "Welcome to china! This was a beautiful county, I think you'll like it. Here is the great Wall ";
str = str. ToLower ();
Dictionary<char, int> dict = new Dictionary<char, int> ();
for (int i = 0; i < str. Length; i++)
//{
if (char. Isletter (Str[i]))
// {
if (Dict. ContainsKey (Str[i]))
// {
dict[str[i]]++;
// }
Else
// {
Dict. ADD (Str[i], 1);
// }
// }
//}
foreach (Keyvaluepair<char, int> kv in dict)
//{
Console.WriteLine ("Character {0}," {1} "appears, KV. Key, KV. Value);
//}
Console.readkey ();
Case: Two (List) set {"A", "B", "C", "D", "E"} and {"D", "E", "F", "G", "H"}, merge the two sets to remove duplicates into one.
List<char> List1 = new list<char> () {' A ', ' B ', ' C ', ' d ', ' e '};
list<char> list2 = new list<char> () {' d ', ' e ', ' f ', ' g ', ' H '};
for (int i = 0; i < List2. Count; i++)
{
If collection 2 is not included in collection 1
if (!list1. Contains (List2[i]))
{
Then add a collection of 2 items to the collection 1
List1. ADD (List2[i]);
}
}
Show the collection 1
for (int i = 0; i < List1. Count; i++)
{
Console.WriteLine (List1[i]);
}
Console.readkey ();
Collection of Exercises
