/*
MergeSort
归并排序
复杂度: n*log(n)
*/
#include<iostream>
using namespace std;
void MergeSort(int a[], int s, int e, int tmp[]);
void Merge(int a[], int s, int m, int e, int tmp[]);
int a[10] = { 13,27,19,2,8,12,2,8,30,89 };
int b[10];
int main()
{
int size = sizeof(a) / sizeof(int);
MergeSort(a, 0, size - 1, b);
for (int k = 0; k < size; k++)
cout << a[k] << " ";
cout << "\n" << endl;
return 0;
}
/*归并排序 s :数组开始 e :数组结尾*/
// 将数组 a 的局部 a[s, m] 和 a[m + 1, e] 合并到 tmp, 并保证 tmp 有序,然后再拷贝回 a[s, m]
// 归并操作时间复杂度: O (e-m+1), 即 O (n)
void MergeSort(int a[], int s, int e, int tmp[])
{
if (s < e) /*当只有一个元素时,递归终止,直接调用merge函数,比较 13和27大小,最后排序*/
{
int m;// cut array into half
m = s + (e - s) / 2;
MergeSort(a, s, m, tmp);/* 排好序的数组 临时放在哪里? */
MergeSort(a, m + 1, e, tmp);
Merge(a, s, m, e, tmp);
}
}
void Merge(int a[],int s, int m, int e, int tmp[])
{
int i = 0;
int p1 = s;
int p2 = m + 1;
while (p1 <= m && p2 <= e)
{
if (a[p1] < a[p2])
{
tmp[i++] = a[p1++];
}
else
{
tmp[i++] = a[p2++];
}
}
while (p1 <= m)
tmp[i++] = a[p1++];
while (p2 <= e)
tmp[i++] = a[p2++];
for (int j = 0; j < e - s + 1; j++)
{
a[s + j] = tmp[j];
}
}
/*
s e m
0 9 4
0 4 2
0 2 1
0 1 0
first:
MergeSort(a, 0, 9, tmp)
13,27,19,2,8
12,2,8,30,89
MergeSort(a, 0, 4, tmp) 13,27,19,2,8
MergeSort(a, 5, 9, tmp) 12,2,8,30,89
second:
MergeSort(a, 0, 4, tmp)
13,27,19
2,8
MergeSort(a, 0, 2, tmp) 13,27,19
MergeSort(a, 3, 4, tmp) 2,8
third:
MergeSort(a, 0, 2, tmp)
13,27
19
MergeSort(a, 0, 1, tmp) 13,27
MergeSort(a, 2, 2, tmp) 19
fourth:
MergeSort(a, 0, 1, tmp)
MergeSort(a, 0, 0, tmp) 13
MergeSort(a, 1, 1, tmp) 27
Merge(a,0,0,1,tmp)
s m e
0 0 1
p1 p2
0 1
1
a[] = {13,27}
assume return to the first cut half , the two half array are sorted
MergeSort(a, 0, 4, tmp) 2,8,13,19,27
MergeSort(a, 5, 9, tmp) 2,8,12,30,89
Merge(a,0,4,9,tmp)
s m e
0 4 9
*/