1.算法原理
2.代码实现
#include <stdio.h>
//printArray打印出数组
void printArray(int a[],int size){
printf("数组为：%d ",a[0]);
for (int i=1;i<size;i++)
{
printf(" %d ",a[i]);
}
printf("\n");
}
//将有二个有序数列a[first...mid]和a[mid...last]合并。
void mergearray(int a[], int first, int mid, int last, int temp[])
{
printArray(a,last);
int i = first, j = mid + 1;
int m = mid, n = last;
int k = 0;
while (i <= m && j <= n)
{
if (a[i] <= a[j])
temp[k++] = a[i++];
else
temp[k++] = a[j++];
}
while (i <= m)
temp[k++] = a[i++];
while (j <= n)
temp[k++] = a[j++];
for (i = 0; i < k; i++)
a[first + i] = temp[i];
}
//递归归并排序
void mergesort(int a[], int first, int last, int temp[])
{
if (first < last)
{
int mid = (first + last) / 2;
mergesort(a, first, mid, temp); //左边有序
mergesort(a, mid + 1, last, temp); //右边有序
mergearray(a, first, mid, last, temp); //再将二个有序数列合并
}
}
bool MergeSort(int a[], int n)
{
int *p = new int[n];
if (p == NULL)
return false;
mergesort(a, 0, n - 1, p);
delete[] p;
return true;
}
void main()
{
int a[10] ={9,8,7,6,5,4,3,2,1,0};
MergeSort(a,10);
printArray(a,10);
}
3.结果
数组为：9
数组为：8 9
数组为：7 8 9 6
数组为：7 8 9 5
数组为：5 6 7 8 9 4
数组为：5 6 7 8 9 3 4
数组为：5 6 7 8 9 2 3 4 1
数组为：5 6 7 8 9 2 3 4 0
数组为：5 6 7 8 9 0 1 2 3
数组为：0 1 2 3 4 5 6 7 8 9
