Solution to 1019

Author: Indigo2233

src/problems/1018.binary-prefix-divisible-by-5.jl

@@ -63,12 +63,12 @@ using LeetCode
 
 function prefixes_div_by5(nums::Vector{Int})
     len = length(nums)
-    res = fill(false, len)    
+    res = fill(false, len)
     cur = 0
     for i in 1:len
         cur = cur * 2 + nums[i]
         cur % 5 == 0 && (res[i] = true)
     end
-    res
+    return res
 end
 ## @lc code=end

src/unresolved/1019.next-greater-node-in-linked-list.jl renamed to src/problems/1019.next-greater-node-in-linked-list.jl

@@ -1,8 +1,8 @@
 # ---
 # title: 1019. Next Greater Node In Linked List
 # id: problem1019
-# author: Tian Jun
-# date: 2020-10-31
+# author: Indigo
+# date: 2022-03-19
 # difficulty: Medium
 # categories: Linked List, Stack
 # link: <https://leetcode.com/problems/next-greater-node-in-linked-list/description/>
@@ -61,5 +61,18 @@
 ## @lc code=start
 using LeetCode
 
-## add your code here:
+function next_larger_nodes(head::ListNode{Int})
+    head = reverse_list(head)
+    stk = Int[]
+    res = Int[]
+    while !isnothing(head)
+        while !isempty(stk) && stk[end] <= val(head)
+            pop!(stk)
+        end
+        pushfirst!(res, isempty(stk) ? 0 : stk[end])
+        push!(stk, val(head))
+        head = next(head)
+    end
+    return res
+end
 ## @lc code=end

src/problems/206.reverse-linked-list.jl

@@ -28,7 +28,8 @@
 ## @lc code=start
 using LeetCode
 
-function reverse_list(head::Union{ListNode,Nothing})::Union{ListNode,Nothing}
+reverse_list(head::Nothing) = head
+function reverse_list(head::ListNode)::ListNode
     new_head = pre_node = nothing
     while !isnothing(head)
         new_head = ListNode(head.val)

test/problems/1019.next-greater-node-in-linked-list.jl

@@ -0,0 +1,6 @@
+@testset "1019.next-greater-node-in-linked-list.jl" begin
+    @test next_larger_nodes(ListNode{Int}([1, 7, 5, 1, 9, 2, 5, 1])) ==
+        [7, 9, 9, 9, 0, 5, 0, 0]
+    @test next_larger_nodes(ListNode{Int}([2, 7, 4, 3, 5])) == [7, 0, 5, 5, 0]
+    @test next_larger_nodes(ListNode{Int}([2, 1, 5])) == [5, 5, 0]
+end