LeetCode/problems/rb/merge-sorted-array.rb
Matej Focko 333866d1bc
chore: split solutions by language
Signed-off-by: Matej Focko <mfocko@redhat.com>
2023-06-02 17:19:02 +02:00

59 lines
1.3 KiB
Ruby

# @param {Integer[]} nums1
# @param {Integer} m
# @param {Integer[]} nums2
# @param {Integer} n
# @return {Void} Do not return anything, modify nums1 in-place instead.
def merge(nums1, m, nums2, n)
# shift numbers to right
i, k = m - 1, nums1.size - 1
while i >= 0 do
nums1[k] = nums1[i]
i -= 1
k -= 1
end
# merge them
i, j, k = nums1.size - m, 0, 0
while i < nums1.size && j < n do
if nums1[i] < nums2[j] then
nums1[k] = nums1[i]
i += 1
else
nums1[k] = nums2[j]
j += 1
end
k += 1
end
while j < n do
nums1[k] = nums2[j]
j += 1
k += 1
end
end
RSpec.describe "merge" do
it "nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 ~> [1,2,2,3,5,6]" do
nums1, m = [1,2,3,0,0,0], 3
nums2, n = [2,5,6], 3
merge(nums1, m, nums2, n)
expect(nums1).to eq([1,2,2,3,5,6])
end
it "nums1 = [0], m = 0, nums2 = [1], n = 1 ~> [1]" do
nums1, m = [1], 1
nums2, n = [], 0
merge(nums1, m, nums2, n)
expect(nums1).to eq([1])
end
it "[3, 3] for 6 is [0, 1]" do
nums1, m = [0], 0
nums2, n = [1], 1
merge(nums1, m, nums2, n)
expect(nums1).to eq([1])
end
end