int[] ary = new int[] { 1, 3, 3, 4, 5, 4 }; var q = from x in ary group x by x into Y select {Y.Key,Y.Count()}; Dictionary<int, int> dict = new Dictionary<int, int>(); foreach (int i in arr) { if (dict.ContainsKey(i)) dict[i]++; else dict.Add(i, 1); }
str.Length
比较下
set的优点之一是不包含重复的元素!
var q = from x in ary
group x by x into Y
select {Y.Key,Y.Count()};
Dictionary<int, int> dict = new Dictionary<int, int>();
foreach (int i in arr)
{
if (dict.ContainsKey(i)) dict[i]++;
else dict.Add(i, 1);
}
我想判断一个二维数组中 某个元素中的某个元素是否相同
string [][] str= {{"姓名","年龄","工作单位","住址"},{"姓名","年龄","工作单位","住址"},{"姓名","年龄","工作单位","住址"}};
// 我就知道如何判断该数组中 姓名是否有重复的
//我现在思维有点小混乱
public bool IsRepeat(List<string> m_list,int start)
{
string index = m_list[start];
for (int i = start+1; i <m_list.Count; i++)
{
if (m_list[i] == index)
return false;
}
IsRepeat(m_list, start + 1);
return true;
}
用泛型类 好操作点,List<> lists=new ...........
bool isTrue= Lists.GroupBy(g => g.姓名).Any(a => a.Count() > 1);
list.Add();
if(list.Contatins())
list.Remove();
//这是我自己想到的可以使用的方法 呵呵 如有不对之处 请指正
List<List<object>> feedback= new List<List<object>>();
List<List<object>> name = new List<List<object>>();
int number = 0;
for (int i = 0; i < feedback.Count; i++)
{
if (feedback[i][7] == null)
{ continue; }
string str = feedback[i][7].ToString();
if (i+1 == feedback.Count)
{
break;
}
if (str == feedback[i + 1][7].ToString())
{
number = i;
}
}
if (number > 1)
{
name.Add(feedback[number]);
name.Add(feedback[number + 1]);
feedback.RemoveAt(number);
feedback.RemoveAt(number);
}