Question

我正试图为下一个问题提出一个解决方案：在OCaml中编写一个函数，该函数具有最长递增子数组的第一个和最后一个索引作为输出。我正在努力修复错误。到目前为止，我已经写了：

示例：

longest_increasing_subarray [|2;4;6;9;1;5;3;5;0;8;9;13;17|];;
- : (8,12)

到目前为止，这是我的代码：

let longest_increasing_subarray p1 =
  let n = Array.length p1
  and beg = ref 0
  and larg = ref(0,0) in
  for i=0 to n-1 do begin
    if i=0 then begin
    end;
    if p1.(i-1)<p1.(i) then
      begin
        if (snd !larg - fst !larg +1)<((i-1)-(!beg+1)) then
          !larg = (!beg,i-1)
            !beg = i end;
    if (i=n-1) then
      begin if (snd !larg-fst !larg +1)<((i)-(!beg)+1) then
          !larg := (!beg,i)
      end;
  end;
  done;
  !larg;;

Answer 1

您的代码和方法有一些问题，让我以问答形式突出它们：

如何为引用分配值
- 正确：
```
x := v
```
- 错误：
```
x = v
!x = v
!x := v
```
如何连续进行两次作业
- 正确：
```
x := v;
y := p
```
- 错误：
```
x := v
y := p
```

如何在条件上分支

正确：

if p1.(i - 1) < p1.(i) 
then update_indices i;
else reset_indices ();

错误：

if p1.(i - 1) < p1.(i) 
then begin update_indices i end
reset_indices ()

如何学习OCaml
- 正确：阅读OCaml manual和other books
- 错：在SO上发帖希望有人能给您代码
如何实现longest_increasing_subarray功能
- 正确：

    Use a recursive function `find_longest` that will have four parameters:
       - `p` - the beginning of the longest sequence;
       - `q` - the end of the longest sequence; 
       - `s` - the beginning of the current working hypothesis
       - `t` - the end of the current working hypothesis;
       - `i` - the array index;
    The longest increasing sequence in the array `a` is defined as
    `find_longest 0 0 0 0 1`. The `find_longest` has the following definiton:
       - if `i >= Array.length a` then the result is 
          - `(p,q)` if `p - q > t - s` 
          - `(s,t)` otherwise
       - else if `a.(i-1) < a.(i)` then the result is `find_longest p q s i (i+1)`
       - else if `q - p < t - s` then the result is `find_longest s t i i (i+1)`
       - else the result is `find_longest p q i i (i+1)`

错误：使用命令式for循环，引用等

Ocaml中最长的递增子数组

1 个答案: