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 }