Question

假设l是一个列表且elem是一个元素，如何在列表elem中返回元素l的最后一次出现？如果元素在l中不存在，也返回-1。我不太明白如何使用递归迭代列表...

let rec getLastOccurence l elem = …

Answer 1

let findi x l = 
  let rec loop i n l = 
    match l with 
    | y::tl -> loop (i+1) (if y = x then i else n) tl 
    | [] -> n 
  in 
  loop 0 (-1) l;;

Answer 2

基本上，您需要两个累加器来跟踪当前索引和元素的最大索引。然后你只需递归到列表的末尾并返回“最大索引”值。

let rindex elem = 
  let rec find_index i max_found = function
    | (x::xs) when x = elem -> find_index (i+1) i xs
    | (_::xs) -> find_index (i+1) max_found xs
    | [] -> max_found
  in find_index 0 (-1);;

这也可以简单地表达为折叠：

let rindex elem ls = 
  let find_index (i, max) elem' = (i+1, if elem' = elem then i else max)
  in snd (fold_left find_index (0, -1) ls);;

Answer 3

这是一种用于在列表中查找整数的尾递归算法：

let find_index elt lst =
  (* Wrap inner function that accepts an accumulator to keep the interface clean *)
  let rec find_it elt acc = function
    | hd :: tl when elt = hd -> acc (* match *)
    | hd :: tl -> find_it elt (acc + 1) tl (* non-match *)
    | _ -> raise Not_found (* end of list *)
  in find_it elt 0 lst (* call inner function with accumulator starting at 0 *)
;;

OCaml中列表中元素的最后出现

3 个答案: