#206

Reverse Linked List

EasyLinked List

Given the head of a singly linked list, reverse the list and return its new head.

A node is ListNode { val int next ListNode? } (an empty list is null).

Example

1 -> 2 -> 3 -> 4 -> 5   =>   5 -> 4 -> 3 -> 2 -> 1

Implement

function ReverseList(head: ListNode?) -> ListNode?

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BAML docs

baml describe

Search the BAML stdlib (260 symbols). Try a type, function, or method name.

let x = 0;

mutable binding

let nums: int[] = [];

typed binding

function Add(a: int, b: int) -> int {
  return a + b;
}

if (x > 0) {
  ...
} else {
  ...
}

while (i < n) {
  i += 1;
}

for (x in nums) {
  ...
}

iterate values

break;   continue;   return v;

nums.length()

nums[i]

read (0-indexed)

nums[i] = v

write

nums.push(x)

nums.pop()

returns T? (optional)

nums.slice(a, b)   nums.reverse()

nums.sort_by_key(k)   nums.sort_by(cmp)

returns a sorted copy (no bare .sort())

let m: map<string, int> = {};

m.set("a", 1);

m.get("a")

returns int? (null if absent)

m.has("a")   m.keys()

m.set(n.to_string(), true);

int keys are unsupported: convert to string

s.length()

number of characters

s.char_at(i)   s.substring(a, b)

s.split(",")   s.to_lower_case()

c.is_alphanumeric()   c == "a"

a / b

integer division

a % b

remainder

x.abs()   x.pow(e)   x.min(y)   x.max(y)

x & 1   x >> 1   x << 2   x ^ y

bitwise ops

n.to_string()

value to string

int.parse(s)

string to int

let node: ListNode? = null;

if (x == null) {
  return 0;
}
// x is non-null below

guard-and-return narrows the type

class ListNode {
  val int
  next ListNode?
}

ListNode { val: 1, next: null }

construct with { }

node.val   node.next