1.最大值与最小值
将一对输入元素相互进行比较,然后把较小的与当前的最小值比较,把较大的与当前最大值比较,这样每两个元素比较3次,最多需要3*n/2次比较可以同时找到一组输入元素中的最大值与最小值。
2.顺序统计量
一个n个元素组成的集合中,第i个顺序统计量是该集合中第i小的元素。中位数是i=(n+1)/2处的元素。
期望为线性时间的选择算法RandomSelect与最坏情况为线性时间的选择算法select如下:
#include<iostream>
using namespace std;
#include<CTime>
#include<vector>
template<class T>
void QuickSort2(T a[], int left, int right)
{
T temp, x = a[right];
int i, j, q;
if (left < right){
i = left - 1;
for (j = left; j < right; j++)
{
if (a[j] <= x)
{
i += 1;
temp = a[j];
a[j] = a[i];
a[i] = temp;
}
}
temp = a[right];
a[right] = a[i + 1];
a[i + 1] = temp;
q = i + 1;
QuickSort2(a, left, q - 1);
QuickSort2(a, q + 1, right);
}
}
int RandomPartition(int a[], int left, int right)
{
srand(time(NULL));
int m = left + rand() % (right - left);
int temp =a[right];
a[right] = a[m];
a[m] = temp;
int x = a[right];
int i, j;
i = left - 1;
for (j = left; j < right; j++)
{
if (a[j] <= x)
{
i += 1;
temp = a[j];
a[j] = a[i];
a[i] = temp;
}
}
temp = a[right];
a[right] = a[i + 1];
a[i + 1] = temp;
return i + 1;
}
int RandomSelect(int a[], int left, int right, int i)
{
if (left == right){ return a[left]; }
int q = RandomPartition(a, left, right);
int k = q - left + 1;
if (i == k)
{
return a[q];
}
else if (i < k){
return RandomSelect(a, left, q - 1, i);
}
else{
return RandomSelect(a, q + 1, right, i - k);
}
}
int Partition(vector<int>&a, int left, int right,int b)
{
int m = 0;
for (int i = left; i <= right; i++){
if (a[i] == b){
m = i;
break;
}
}
int temp = a[right];
a[right] = a[m];
a[m] = temp;
int x = a[right];
int i, j;
i = left - 1;
for (j = left; j < right; j++)
{
if (a[j] <= x)
{
i += 1;
temp = a[j];
a[j] = a[i];
a[i] = temp;
}
}
temp = a[right];
a[right] = a[i + 1];
a[i + 1] = temp;
return i + 1;
}
//原数组使用vector是因为寻找中位数的中位数时的数组不固定,但也需要调用select函数
int select(vector<int>&a, int left, int right, int dst)
{
int n = right - left + 1;
if (n == 1){ return a[left]; }
vector<int> temp;
vector<int> b;
for (int i = left; i <right+1; i += 5)
{
if (right+1 - i < 5){ break; }
for (int j = i; j - i < 5; j++)
{
b.push_back(a[j]);//不能再原址进行排序,否则可能会改变分割后的数组
}
//插入排序
for (int j = 1; j < 5; j++)
{
int t = b[j];
int m = j - 1;
while (m >= 0 && b[m]>t)
{
b[m + 1] = b[m];
m--;
}
b[m + 1] = t;
}
temp.push_back(b[2]);
b.clear();
}
if (n%5!= 0){
for (int i =left+ (n / 5) * 5 ; i < right+1; i++)
{
b.push_back(a[i]);
}
//插入排序
for (int i = 1; i<b.size();i++)
{
int t = b[i];
int j = i - 1;
while (j >=0 && b[j]>t)
{
b[j + 1] = b[j];
j--;
}
b[j + 1] = t;
}
temp.push_back(b[(b.size()+1)/2-1]);
b.clear();
}
int q = Partition(a, left, right, select(temp, 0, temp.size()-1, (temp.size()+1) / 2));
int k = q - left + 1;
if (dst == k)
{
return a[q];
}
else if (dst < k){
return select(a, left, q - 1, dst);
}
else{
return select(a, q + 1, right, dst - k);
}
}
int main()
{
int a[22];
vector<int> m;
srand(time(NULL));
cout << "原数组为:" << endl;
for (int i = 0; i < 22; i++)
{
a[i] = rand() % 100;
m.push_back(a[i]);
cout << a[i] << " ";
}
cout << endl;
int num =12;
int b=RandomSelect(a, 0, 21, num);
cout << "方法一:第"<<num<<"小的数为:"<<b << endl;
b = select(m, 0, 21, num);
cout << "方法二:第" << num << "小的数为:" << b << endl;
QuickSort2<int>(a, 0, 21);
cout << "快速排序后为:" << endl;
for (int i = 0; i < 22; i++)
{
cout << a[i] << " ";
}
cout << endl;
}