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### Author Topic: Permutation Algorithm in C  (Read 1119 times)

#### kathir

• Sr. Member
• Posts: 283
##### Permutation Algorithm in C
« on: June 08, 2011, 04:17:04 am »
I have implemented permutation algorithm on Turbo C, Visual C++, C# and Java.

Turbo C/C++ programmers: http://www.softwareandfinance.com/Turbo_CPP/MRP.html

Visual C++ programmers: http://www.softwareandfinance.com/Visual_CPP/MRP.html

C# programmers: http://www.softwareandfinance.com/CSharp/MRP.html

Java programmers: http://www.softwareandfinance.com/Java/MRP.html

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#include <stdio.h>
#include <string.h>

void sortchar(char *buffer, int len)
{
int i, j;
for(i = 1; i < len; i++)
{
for(j = 0; j < len - i; j++)
{
if(buffer[j] > buffer[j + 1])
{
char temp = buffer[j];
buffer[j] = buffer[j + 1];
buffer[j + 1] = temp;
}
}
}
}

int NextPermuation(char *p, int len)
{
int i,j,k;
int anchor, count;
char *tempptr;
char ch, newchar;

for(k = len - 1; k > 0; k--)
{
if(p[k - 1] >= p[k])
continue;
else
{
if(k <= len - 3)
{
newchar = p[k-1];
anchor = -1;
for(j = len - 1; j >= k; j--)
{
if(newchar < p[j])
{
anchor = j;
break;
}
}
if(anchor == -1)
return 0;
ch = p[k-1];
p[k-1] = p[anchor];
p[anchor] = ch;
// sort last remaining chars
sortchar(p+k,len - k);
return 1;
}
else
{
tempptr = &p[len-3];
count = 3;
for(i = count - 1; i > 0; i--)
{
if(tempptr[i - 1] >= tempptr)
continue;
else
{
if(i <= count - 2)
{
if(tempptr[i+1] > tempptr[i-1])
{
ch = tempptr[i+1];
tempptr[i+1] = tempptr;
tempptr = tempptr[i-1];
tempptr[i-1] = ch;
}
else
{
ch = tempptr[i-1];
tempptr[i-1] = tempptr;
tempptr = tempptr[i+1];
tempptr[i+1] = ch;
}
}
else
{
ch = tempptr;
tempptr = tempptr[i-1];
tempptr[i-1] = ch;
}
return 1;
}
}
return 0;
}
}
}
return 0;
}

int main(int argc, char* argv[])
{
char buf[32];
int ret;
int count = 0;

printf("Enter a string to find permutation:");
scanf("%s", buf);
//sortchar(buf, strlen(buf));

while(1)
{
printf("%s\n", buf);
count++;
ret = NextPermuation(buf, strlen(buf));
if(ret == 0)
break;
}

printf("\n\nCount: %d\n\n\n", count);
return 0;
}

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