kkaushi
BAN USER
Are you sure it's O(n).
for(i = 0; i < len; i++) {
int new_idx = get_index(i, N);
while(new_idx < i) {
new_idx = get_index(new_idx, N);
}
printf("i: %d; new_idx: %d\n", i, new_idx);
swap(&s[i], &s[new_idx]);
}
What about the while loop inside for. Does it always takes constant time?
I think it's O(n^2) as inner while loop is dependent on n or it's O(n)
a[-1] points to i because when the line
int a[3],i;
is encountered during execution, space is allocated in stack equal to space of 3 ints (for array a) + space for 1 integer (ie i). Let the adress of a[0] be 0x00000008 (for simplification and assuming 32 bit int size). Then a[1] would be at address 0x0000000c and a[2] would have address 0x000000010. The address of i would be 4 bytes before a[0]'s address in stack ie 0x00000006 because the space is allocated in Right To Left order ie i first, followed by a[3].
- kkaushi August 03, 2014- 0
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That won't work as you need to interchange left/right child of all nodes in subtree of root->left and root->right.
- kkaushi August 04, 2014