How can I check two “pointers” meet each other when going through a list?

问题

For example, I have a list [1; 2; 1; 3; 2; 4; 1; 5; 6; 8;...]. The list has duplicate elements.

Then we go through the list by two virtual pointers.

One pointer a goes from the beginning and has speed of 2.

The other pointer b goes from somewhere in the middle of list and has speed of 1.

So how can I check whether a meeting with b?

For example:

a starts from [1; 2; 1; 3; 2; 4; 1; 5; 6; 8;...]

b starts from index 1, which is [2; 1; 3; 2; 4; 1; 5; 6; 8;...]

So after one time move, both a and b will arrive at index 2 and they meet.

How can I check whether meet or not by the element only, without info of index?

We can just compare the value of the two elements, as there might be duplicates inside.

In term of code:

let l = [1; 2; 1; 3; 2; 4; 1; 5; 6; 8;...]

let find_middle_start l n = 
  let rec aux i = function
    | [] -> []
    | _::tl when i >= n -> tl
    | _::tl -> aux (i+1) tl
  in
  aux 0 l

let parallel_travel l =
  Random.self_init();
  let b_start = find_middle_start l (Random.int (List.length l)) in
  let rec go = function
    | [], _::_ | _::_, [] | _::[], _::_ -> print_endline "Never meet"
    | x::x1::xs, y::ys ->
      if check_meet x y then print_endline "Meet"
      else go xs ys
  in
  go l b_start

How can I implement check_meet? i can't just do x == y right?

回答1:

Bind the lists, and compare them by identity (with ==).

| ((_::_::xs) as a)), ((_::ys) as b) ->
  if a == b then ...
  else ...

Note that if you are trying to determine whether a list shares a tail with another list, this method is not sufficient (the slow pointer can reach the end before the fast pointer catches up with it).

来源：https://stackoverflow.com/questions/29121354/how-can-i-check-two-pointers-meet-each-other-when-going-through-a-list

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