对了,那个第四个问题有O(N)的解法,是快速选择的思想
下面是我实现的代码
#include <iostream>
using namespace std;
#define MAX_SIZE 2001
//int a[2001];
void swap(int *a, int *b)
{
int tmp = *a;
*a = *b;
*b = tmp;
}
int partition(int a[], int low, int high)
{
int privotKey = a[low]; //基准元素
while (low < high){ //从表的两端交替地向中间扫描
while (low < high && a[high] >= privotKey) --high; //从high 所指位置向前搜索,至多到low+1 位置。将比基准元素小的交换到低端
swap(&a[low], &a[high]);
while (low < high && a[low] <= privotKey) ++low;
swap(&a[low], &a[high]);
}
//a[low] = privotKey;
return low;
}
void Top100(int a[], int k,int start,int end){
int i = start;
int j = end - 1;
int index = partition(a, i, j);
while (index!=end-k-1) //数组后100
{
if (index<end - k - 1)
{
i = ++index;
index = partition(a, i, j);
}
else
{
j = index;
index = partition(a, i, j);
}
}
}
int main()
{
int a[] = { 1111, 22222, 3333, 4, 5, 6, 7, 8,9 ,10};
Top100(a, 3, 0, 10);
for (int i = 0; i <10 ; i++)
{
cout << a[i]<<endl;
}
return 0;
}