Input :
A[4] = {0,4,-1,1000} - Actual Array
P[4] = {1,0,3,2} - Order to be r开发者_如何学JAVAeshuffled
Output:
A[4] = {4,0,1000,-1}
Condition : Don't use an additional array as memory. Can use an extra variable or two.
Problem : I have the below program in C++, but this fails for certain inputs of array P.
#include<iostream>
using namespace std;
void swap(int *a_r,int *r)
{
int temp = *r;
*r = *a_r;
*a_r = temp;
}
int main()
{
int A[4] = {0,4,-1,1000};
int P[4] = {3,0,1,2};
int value = A[0] , dest = P[0];
for(int i=0; i<4;i++)
{
swap(&A[dest],&value);
dest = P[dest];
}
for(int i=0;i<4;i++)
cout<<A[i]<<" ";
}
Since you've got a spare array called P kicking around, and there isn't anything in the question as quoted that stipulates it must be treated as a constant array, you could do:
for (i = 0; i < 4; i++)
P[i] = A[P[i]];
for (i = 0; i < 4; i++)
A[i] = P[i];
If you're not allowed to modify P, then you have to work a lot harder (and you also have to work a lot harder if the data type of A is not the same as, or compatible with, the type of P).
However, I fear this is a sleight-of-hand trick that doesn't really answer the question; it gets the job done, but doesn't answer the question.
First of all, I really like Jonathan's solution, but I feel like I can add some interesting ideas too.
The main observation is that array P
consists of several loops.
Let's consider p = {1, 4, 3, 2, 0, 5}
. There're three loops: 0 => 1 => 4 => 0
, 2 => 3 => 2
and 5 => 5
. And to replace variables alongside one loop we need no additional memory at all. We just go through it like this
do {
a[i] = a[p[i]];
i = p[i];
} while (i != first_i);
(The last element needs to be taken special care of, though.) The full working version:
for (int i = 0; i < n; ++i) {
if (p[i] < 0) {
// been at index 'i' already
continue;
}
// new loop found
int j = i;
int first_value = a[i]; // to be put in last position in the chain
int prev_j; // we always store previous 'j' index
do {
a[j] = a[p[j]];
prev_j = j;
j = p[j]; // move to next 'j'
p[prev_j] = -1; // mark element as processed
} while (i != j);
a[prev_j] = first_value;
}
The only problem with my solution is that it uses p
array to mark element as 'processed'. Some interviewers may consider it ok, others - not, depending on the solution they have in mind.
int _tmain(int argc, _TCHAR* argv[])
{
A[4] = {0,4,-1,1000}
P[4] = {1,0,3,2}
int temp = arr2[0];
for(int i=0;i<4;i++)
{
for(temp = P[i];i<3;temp = P[temp])
{
if(temp >= i)
{
int data;
data = A[i];
A[i] = A[temp];
A[temp] = data;
break;
}
}
}
_getch();
return 1;
}
int _tmain(int argc, _TCHAR* argv[])
{
A[4] = {0,4,-1,1000}
P[4] = {1,0,3,2}
int temp = arr2[0];
for(int i=0;i<4;i++)
{
for(temp = P[i];i<3;temp = P[temp])
{
if(temp >= i)
{
int data;
data = A[i];
A[i] = A[temp];
A[temp] = data;
break;
}
}
}
_getch();
return 1;
}
Slightly modified to calculate dest value
int main()
{
int A[4] = {0,4,-1,1000};
int P[4] = {3,0,1,2};
int value = A[0], dest = P[0];
for(int i=0; i<4-1;i++)
{
int count=0;
dest = P[i];
while(dest<i){ //if P[i] points to lower value, it got swapped with some other position.
dest = P[dest];
}
swap(&A[dest],&A[i]);
}
for(int i=0;i<4;i++)
cout<<A[i]<<" ";
cout<<"\n";
}
Could argue that it goes against the spirit of the question - but can use multiple stack instances (from a run-time perspective) of a single local variable (code perspective). Being allowed to mutate P is just as uncertain, so, FWIW - a recursive answer...
template <int N>
void shuffle(int (&a)[N], int p[], int i = -1)
{
if (++i < N)
{
int result = a[p[i]];
shuffle(a, p, i);
a[i] = result;
}
}
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