#include<iostream>
using namespace std;
void merge(int num[], int tempnum[], int L, int R);
void mergesourt(int num[], int tempnum[], int L, int R);
int _tmain(int argc, _TCHAR* argv[])
{
int a[6] = { 1, 6, 5, 9, 8, 7 };
int b[6];
mergesourt(a, b, 0, 5);
for (int i = 0; i < 6; i++)
{
cout << a[i] << " ";
}
cout << endl;
return 0;
}
void merge(int num[], int tempnum[], int L, int R)
{
int i = 0,j=L,k=R;
int mid = R - ((R - L) >> 1);
while (j <= mid&&k >= mid + 1)
{
tempnum[i++] = (num[j] < num[k]) ? num[j++] : num[k++];
}
while (j <= mid)
{
tempnum[i++] = num[j++];
}
while (k >= mid + 1)
{
tempnum[i++] = num[k++];
}
for (i = L; i <= R; i++)
{
num[i] = tempnum[i];
}
}
void mergesourt(int num[], int tempnum[], int L, int R)
{
int mid = R - ((R - L) >> 1);
if (L < R)
{
mergesourt(num, tempnum, L, mid);
mergesourt(num,tempnum, mid + 1, R);
merge(num, tempnum, L, R);
}
}
这个mergesourt函数啊,当L=0,R=1的时候,mid是1,因此会陷入无限递归
(为何不使用L+(R-L)/2呢?)
另外,merge函数中为什么是k>=mid+1?这样循环就不会停啊?
建议给代码加上高亮
– dudu 6年前