快速排序无限循环数值配方
Quick Sort infinite loop Numerical Recipies
我试图从MATLAB导出一些代码到MEX文件(a.la C)。对于我试图导出的代码段,我需要实现一种可以对2D数组进行排序的算法。打开一份数字食谱,我发现了基于索引的排序。我已经用零索引实现了这个函数,以一种简单的方式,稍后我将对其进行简化。目前的问题是,当我运行我的函数时,它将在传递某些随机数集时进入无限循环。例如:
0.8147,
0.9058,
0.127,
0.9134,
0.6324,
0.0975,
0.2785,
0.5469,
0.9575,
0.9649,
0.1576,
0.9706,
0.9572,
0.4854,
0.8003,
0.1419,
0.4218,
0.9157,
0.7922,
0.9595
这些数字都不重复。下面是我的C代码版本的数值配方。有什么问题吗?我在这上面花了3个小时。该算法为表现良好的随机数集提供适当的索引。
#define SWAP(a,b) itemp=(a);(a)=(b);(b)=itemp;
#define M 7
#define NSTACK 50
long* IndexSort(unsigned long int vectorLength, double* column)
{
unsigned long i, indxt, ir = vectorLength, itemp,j, k, l = 1;
int jstack = 0, *istack;
long *indx;
double a;
istack = (int*)malloc(NSTACK*sizeof(int));
//initalize output
indx = (long*)malloc(vectorLength*sizeof(long));
for (j = 0; j < vectorLength; j++)
{
indx[j] = j;
}
//
while (true)
{
if (ir - l < M)
{
for (j = l+1; j <= ir;j++)
{
indxt = indx[j-1];
a = column[indxt];
for (i = j-1; i >= 1; i--)
{
if (column[indx[i - 1]] <= a)
{
break;
}
indx[i + 1 - 1] = indx[i - 1];
}
indx[i + 1 - 1] = indxt;
}
if (jstack == 0)
{
break;
}
ir = istack[jstack--];
l = istack[jstack--];
}
else
{
k = (l + ir) >> 1;
SWAP(indx[k - 1], indx[l + 1 - 1])
if (column[indx[l + 1 - 1]] > column[indx[ir - 1]]){
SWAP(indx[l + 1 - 1], indx[ir - 1])
}
if (column[indx[l - 1]] > column[indx[ir - 1]]){
SWAP(indx[l - 1], indx[ir - 1])
}
if (column[indx[l + 1 - 1]] > column[indx[l - 1]]){
SWAP(indx[l + 1 - 1], indx[l - 1])
}
i = l + 1;
j = ir;
indxt = indx[l - 1];
a = column[indxt];
while (true)
{
do i++; while (column[indx[i - 1]] < a);
do j--; while (column[indx[j - 1]] > a);
if (j < i) break;
SWAP(indx[i - 1], indx[j - 1])
}
indx[l - 1] = indx[j - 1];
indx[j - 1] = indxt;
jstack += 2;
if (jstack > NSTACK) error("NSTACK too small for IndexSort");
if (ir - i + 1 > j - 1){
istack[jstack] = ir;
istack[jstack - 1] = l;
ir = j - 1;
}
else{
istack[jstack] = j - 1;
istack[jstack - 1] = l;
l = i;
}
}
}
free(istack);
return indx;
}
编辑1:我在顶部添加了#define。作为对第一波评论的回应:
- 这个函数调用产生一个排序索引数组。我将使用这些值对2D数组进行排序。对不起,我应该说得更清楚一点。
- 至于使用调试器,目前是不可能的。该函数是从一个MEX包装器调用的。我将编写一个main函数来研究这个问题。
编辑2:
- 按照建议移动free命令,仍然没有变化。
- 更新问题说明我在论坛上看到问题,大多数人都崩溃了。我的问题实际上是一个无限循环,我不知道它在哪里开始。
提前感谢你的帮助,
如果目标是对给定数组中的索引进行排序,那么使用c++ 11, std::sort
和lambdas的以下代码应该可以完成这项工作:
c++ 11例
#include <algorithm>
#include <iostream>
#include <array>
long* IndexSort(unsigned long int vectorLength, double* column)
{
long *index = new long[vectorLength](); // I really don't recommend this line
long n = 0;
std::generate( index, index + vectorLength, [&] { return n++; });
std::sort(index, index + vectorLength, [&](long v1, long v2)
{ return column[v1] < column[v2]; });
return index;
}
using namespace std;
int main()
{
std::array<double,20> testData =
{0.8147,0.9058,0.127,0.9134,0.6324,0.0975,0.2785,
0.5469,0.9575,0.9649,0.1576,0.9706,0.9572,0.4854,
0.8003,0.1419,0.4218,0.9157,0.7922,0.9595};
long *indices = IndexSort(testData.size(), &testData[0]);
for (size_t i = 0; i < testData.size(); ++i )
cout << testData[indices[i]] << " has an index of " << indices[i] << "n";
delete [] indices;
}
实例:http://ideone.com/EsOFKt
注意,我们所做的只是构建一个排序的索引数组,并使用传递给lambda的两个索引比较原始双向量。
注意,在开始时为索引分配内存是不必要的,因为std::vector<long>
是通常的方法。然而,我试图模仿你的原始代码要求返回一个指针到动态分配的内存。
c++ 0x, 98 example
如果您使用的是c++ 11之前的编译器,您可以将代码更改为:
#include <algorithm>
#include <iostream>
struct IndexSorter
{
double *m_array;
IndexSorter(double *oArray) : m_array(oArray) {}
bool operator()(long v1, long v2) const { return m_array[v1] < m_array[v2]; }
};
long* IndexSort(unsigned long int vectorLength, double* column)
{
long *index = new long[vectorLength](); // I really don't recommend this line
for (unsigned long i = 0; i < vectorLength; ++i)
index[i] = i;
std::sort(index, index + vectorLength, IndexSorter(column));
return index;
}
using namespace std;
int main()
{
double testData[] = {0.8147,0.9058,0.127,0.9134,0.6324,0.0975,0.2785,
0.5469,0.9575,0.9649,0.1576,0.9706,0.9572,0.4854,
0.8003,0.1419,0.4218,0.9157,0.7922,0.9595};
unsigned long testSize = sizeof(testData) / sizeof(testData[0]);
long *indices = IndexSort(testSize, &testData[0]);
for (unsigned long i = 0; i < testSize; ++i )
cout << testData[indices[i]] << " has an index of " << indices[i] << "n";
delete [] indices;
}
实例:http://ideone.com/nhMwSs
<
C例子/strong>
由于这个问题也被标记为C
,这里是C
实现:
#include <stdlib.h>
#include <stdio.h>
static double *myArray;
int compareIndices(const void* v1, const void *v2)
{
long val1 = *(long *)v1;
long val2 = *(long *)v2;
if (myArray[val1] < myArray[val2])
return -1;
else
if ( myArray[val1] > myArray[val2])
return 1;
return 0;
}
long* IndexSort(unsigned long int vectorLength, double* column)
{
myArray = column;
long *index = malloc(vectorLength * sizeof(long));
if ( index )
{
unsigned long i;
for (i = 0; i < vectorLength; ++i)
index[i] = i;
qsort(index, vectorLength, sizeof(long), compareIndices);
return index;
}
return 0;
}
int main()
{
double testData[] = { 0.8147, 0.9058, 0.127, 0.9134, 0.6324, 0.0975, 0.2785,
0.5469, 0.9575, 0.9649, 0.1576, 0.9706, 0.9572, 0.4854,
0.8003, 0.1419, 0.4218, 0.9157, 0.7922, 0.9595 };
unsigned long numItems = sizeof(testData) / sizeof(testData[0]);
long *indices = IndexSort(numItems, testData);
if ( indices )
{
unsigned long i;
for (i = 0; i < numItems; ++i)
printf("%lf has an index of %ldn", testData[indices[i]], indices[i]);
free(indices);
}
}
实例:http://ideone.com/8SZH1m
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