选第k小元素：特定分治策略

1. 问题

在一个数组中找到第k小的元素并输出

2. 解析

 

 

 

 

3. 设计

int select(int a[], int low, int high, int k)

{

int middle;

 

middle = split(a, low, high);

if (middle == k)

return a[k];

else if (middle < k)

return select(a, middle + 1, high, k);

else   

return select(a, low, middle - 1, k);

}

4. 分析

时间复杂度为o(n)

5. 源码

 

#include <stdio.h>

#include <algorithm>

#include <math.h>

using namespace std;

 

int partition(int A[], int low, int high)

{

int pivot = A[low];   

while (low < high){

while (low < high && A[high] >= pivot){

--high;

}

A[low] = A[high];  

while (low < high && A[low] <= pivot){

++low;

}

A[high] = A[low];

}

A[low] = pivot;

return low;

}

 

int r = 5;

//A[low..high]

int select_rank_k(int A[], int low, int high, int k)

{

int r_group = ceil((high - low + 1)*1.0 / r);

//计算每个分组中值，存于A[]最前面

for (int i = 1; i <= r_group; ++i) {

sort(&A[low + (i - 1)*r], &A[(low + i*r - 1) > high ? high : (low + i*r - 1)]);

swap(A[low + i - 1], A[low + (i-1)*r + r / 2]);

}

sort(&A[low], &A[low + r_group]);

swap(A[low], A[r_group / 2]);

int cur = partition(A, low, high);

if (cur == k-1){

return A[cur];

}

else if (cur < k){

return select_rank_k(A, cur + 1, high, k);

}

else{

return select_rank_k(A, low, cur - 1, k);

}

}

 

int main(void)

{

int A[15] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 };

printf("%d\n", select_rank_k(A, 0, 3, 2));

return 0;

 

}