add probelm 1245 (#171)

Author: Moelf

src/problems/1051.height-checker.jl

@@ -64,6 +64,6 @@ using LeetCode
 
 function height_checker(heights::Vector{Int})
     sorted = sort(heights)
-    count(i -> heights[i] != sorted[i], 1:length(heights)) 
+    return count(i -> heights[i] != sorted[i], 1:length(heights))
 end
 ## @lc code=end

src/problems/1054.distant-barcodes.jl

@@ -58,6 +58,6 @@ function rearrange_barcodes(barcodes::Vector{Int})
             idx > len && (idx = 2)
         end
     end
-    res
+    return res
 end
 ## @lc code=end

src/problems/1245.tree-diameter.jl

@@ -0,0 +1,81 @@
+# ---
+# title: 1245. Tree Diameter
+# id: problem1245
+# author: Jerry Ling
+# date: 2023-06-08
+# difficulty: Medium
+# categories: Tree
+# link: <https://leetcode.com/problems/tree-diameter/>
+# hidden: true
+# ---
+# 
+# Given an undirected tree, return its diameter: the number of edges in a longest path in that tree.
+#
+# The tree is given as an array of edges where edges[i] = [u, v] is a bidirectional edge between nodes u and v.  Each node has labels in the set {0, 1, ..., edges.length}.
+# 
+# 
+# **Example 1:**
+# 
+# Input: edges = [[0,1],[0,2]]
+# Output: 2
+# Explanation: 
+# A longest path of the tree is the path 1 - 0 - 2.
+#     
+# 
+# **Example 2:**
+# 
+# Input: edges = [[0,1],[1,2],[2,3],[1,4],[4,5]]
+# Output: 4
+# Explanation: 
+# A longest path of the tree is the path 3 - 2 - 1 - 4 - 5.     
+# 
+# 
+# 
+# **Constraints:**
+# 
+#   * 0 <= edges.length < 10^4
+#   * edges[i][0] != edges[i][1]
+#   * 0 <= edges[i][j] <= edges.length
+# 
+# 
+## @lc code=start
+using LeetCode
+
+function tree_diameter(edges::Vector{Vector{Int}})
+    n = length(edges)
+    neighbors = Dict(i => Int[] for i in 0:n)
+    for edge in edges
+        u, v = edge
+        push!(neighbors[u], v)
+        push!(neighbors[v], u)
+    end
+
+    function depth_first(node, neighbors)
+        # (current length, current node)
+        longest = (0, node)
+
+        # each tuple is (current length, current node, previous node)
+        todo = [(longest..., -1)]
+        while !isempty(todo)
+            len, this, prev = pop!(todo)
+            if len > longest[1]
+                longest = (len, this)
+            end
+
+            for next in neighbors[this]
+                next == prev && continue
+                push!(todo, (len + 1, next, this))
+            end
+        end
+
+        return longest
+    end
+
+    # we find one "end" of the eventual chain by doing depth first once
+    # and a second time starting from the known end
+    _, one_end = depth_first(0, neighbors)
+    diameter, _ = depth_first(one_end, neighbors)
+
+    return diameter
+end
+## @lc code=end

src/problems/1513.number-of-substrings-with-only-1s.jl

@@ -67,6 +67,6 @@ using LeetCode
 
 function num_sub(s::String)
     ss = split(s, '0')
-    sum(binomial(length(subs) + 1, 2) for subs in ss)    
+    return sum(binomial(length(subs) + 1, 2) for subs in ss)
 end
 ## @lc code=end

src/problems/904.fruit-into-baskets.jl

@@ -97,6 +97,6 @@ function total_fruit(fruits::Vector{Int})
         end
         res = max(res, i - j + 1)
     end
-    res
+    return res
 end
 ## @lc code=end

test/problems/1245.tree-diameter.jl

@@ -0,0 +1,3 @@
+@testset "1245.tree-diameter" begin
+    @test tree_diameter([[0,1],[1,2],[2,3],[1,4],[4,5]]) == 4
+end

test/problems/886.possible-bipartition.jl

@@ -1,5 +1,5 @@
 @testset "886.possible-bipartition.jl" begin
-    @test possible_bipartition(4, [[1,2],[1,3],[2,4]])
-    @test !possible_bipartition(3, [[1,2],[1,3],[2,3]])
-    @test !possible_bipartition(5, [[1,2],[2,3],[3,4],[4,5],[1,5]])
+    @test possible_bipartition(4, [[1, 2], [1, 3], [2, 4]])
+    @test !possible_bipartition(3, [[1, 2], [1, 3], [2, 3]])
+    @test !possible_bipartition(5, [[1, 2], [2, 3], [3, 4], [4, 5], [1, 5]])
 end

test/problems/904.fruit-into-baskets.jl

@@ -1,6 +1,6 @@
 @testset "904.fruit-into-baskets.jl" begin
-    @test total_fruit([1,2,1]) == 3
-    @test total_fruit([0,1,2,2]) == 3
-    @test total_fruit([1,2,3,2,2]) == 4
-    @test total_fruit([3,3,3,1,2,1,1,2,3,3,4]) == 5
+    @test total_fruit([1, 2, 1]) == 3
+    @test total_fruit([0, 1, 2, 2]) == 3
+    @test total_fruit([1, 2, 3, 2, 2]) == 4
+    @test total_fruit([3, 3, 3, 1, 2, 1, 1, 2, 3, 3, 4]) == 5
 end