输入n个数，找出其中最小的k个数。例如输入4,5,1,6,2,7,3,8 这8个数，则最小的4个数是1,2,3,4.
解法一：O（n）的算法，只有当我们可以修改输入数组时可用
1 int Partitoin(int* input, int low, int high)
2 {
3 if (input == NULL || low < 0 || high < 0 || low > high)
4 {
5 return -1;
6 }
7 int temp = input[low];
8 while(low < high)
9 {
10 while (high>low && input[high] >= temp)
11 {
12 high--;
13 }
14 input[low] = input[high];
15 while (low<high && input[low] <= temp)
16 {
17 low++ ;
18 }
19 input[high] = input[low];
20 }
21 input[low] = temp ;
22
23 return low ;
24 }
25
26 void GetLeastNumbers(int* input , int length, int k)
27 {
28 if (input == NULL || length < 1 || k < 0 || k >length)
29 {
30 return;
31 }
32 int low = 0 ;
33 int high = length - 1 ;
34 int index = Partitoin(input, low, high);
35 while (index != (k - 1))
36 {
37 if (index < (k - 1))
38 {
39 low = index + 1;
40 index = Partitoin(input, low, high);
41 }
42 else
43 {
44 high = index - 1 ;
45 index = Partitoin(input, low, high);
46 }
47 }
48 for (int i = 0 ; i < k ; i++)
49 {
50 cout<<input[i]<<" ";
51 }
52 }
解法二：O（nlogk）的算法，特别适合处理海量数据
1 typedef multiset<int , greater<int> > intSet ;
2 typedef intSet::iterator setIterator ;
3 void GetLeastNumbers(vector<int>& data , intSet& leastNumbers, int k)
4 {
5 if (k < 1 || data.size() < k)
6 {
7 return;
8 }
9
10 vector<int>::iterator ite = data.begin();
11 for (;ite != data.end() ; ite++)
12 {
13 if ( (leastNumbers.size()) < k)
14 {
15 leastNumbers.insert(*ite);
16 }
17 else
18 {
19 setIterator iterGreatest = leastNumbers.begin() ;
20 if (*ite < *iterGreatest)
21 {
22 leastNumbers.erase(iterGreatest);
23 leastNumbers.insert(*ite);
24 }
25 }
26
27 }
28 }
时间： 2024-10-05 00:37:18