从长度为n的整形数组中返回m个最大数(不许使用排序)

 1 public static int[] GetTheManyMaxValues(int[] arr, int number)
 2 {
 3     if (arr == null || number <= 0 || arr.Length < number) return null;
 4     List<int> lst = new List<int>();
 5     for (int i = 0; i < number; i++)
 6     {
 7         lst.Add(arr[i]);
 8     }
 9     for (int i = number; i < arr.Length; i++)
10     {
11         int min = GetTheMinValue(lst);
12         if (min < arr[i])
13         {
14             lst.Remove(min);
15             lst.Add(arr[i]);
16         }
17     }
18     return lst.ToArray();
19 }

 1 public static int GetTheMinValue(List<int> lst)
 2 {
 3     int min = lst[0];
 4     foreach (int number in lst)
 5     {
 6         if (min > number)
 7         {
 8             min = number;
 9         }
10     }
11     return min;
12 }