Check if two linked lists merge. If so, where?

Question

This question may be old, but I couldn\'t think of an answer.

Say, there are two lists of different lengths, merging at a point; how do we know wher

Answer 1

I have tested a merge case on my FC9 x86_64, and print every node address as shown below:

Head A 0x7fffb2f3c4b0
0x214f010
0x214f030
0x214f050
0x214f070
0x214f090
0x214f0f0
0x214f110
0x214f130
0x214f150
0x214f170


Head B 0x7fffb2f3c4a0
0x214f0b0
0x214f0d0
0x214f0f0
0x214f110
0x214f130
0x214f150
0x214f170

Note becase I had aligned the node structure, so when malloc() a node, the address is aligned w/ 16 bytes, see the least 4 bits. The least bits are 0s, i.e., 0x0 or 000b. So if your are in the same special case (aligned node address) too, you can use these least 4 bits. For example when travel both lists from head to tail, set 1 or 2 of the 4 bits of the visiting node address, that is, set a flag;

next_node = node->next;
node = (struct node*)((unsigned long)node | 0x1UL);

Note above flags won't affect the real node address but only your SAVED node pointer value.

Once found somebody had set the flag bit(s), then the first found node should be the merge point. after done, you'd restore the node address by clear the flag bits you had set. while an important thing is that you should be careful when iterate (e.g. node = node->next) to do clean. remember you had set flag bits, so do this way

real_node = (struct node*)((unsigned long)node) & ~0x1UL);
real_node = real_node->next;
node = real_node;

Because this proposal will restore the modified node addresses, it could be considered as "no modification".