#define Max 2147483647
#include <iostream>
#include <cstdlib>
using namespace std;
void Merge(int *A, int p, int q, int r)
{
int i, j;
int n1 = q - p + 1;
int n2 = r - q;
//归并堆
int *L = new int[n1 + 1];
//归并堆
int *R = new int[n2 + 1];
for (i = 0; i < n1; ++i)
{
L[i] = A[p + i - 1];
}
for (j = 0; j < n2; ++j)
{
R[j] = A[q + j];
}
L[n1] = Max;
R[n2] = Max;
i = 0;
j = 0;
for (int k = p - 1; k < r; ++k)
{
if (L[i] <= R[j])
{
A[k] = L[i++];
}
else
{
A[k] = R[j++];
}
}
}
void MergeSort(int *A, int p, int r)
{
if (p < r)
{
int q = (p + r) / 2;
MergeSort(A, p, q);
MergeSort(A, q + 1, r);
Merge(A, p, q, r);
}
}
int main(void)
{
int A[7] = { 18, 45, 36, 30, 92, 72, 25 };
MergeSort(A, 1, 7);
for (int i = 0; i < 7; ++i)
{
cout << A[i] << " ";
}
cout << endl;
system("pause");
return 0;
}

分析:T(n) = c 若n=1
=2T(n/2)+cn 若n>1
解得:T(n) = 2^(log(2)n)T(1) + (2^0+2^1+···+2^(log(2)n - 1))cn
=2^(log(2)n)c + (2^0+2^1+···+2^(log(2)n - 1))cn
总代价为cnlgn+cn,它就是O(nlgn)。
原文:http://www.cnblogs.com/mubu/p/5786033.html