在红宝石中重叠切片

Question

鉴于我有数组[1,2,3,4,5,6,7,8,9,10],一次抓取2个元素的最短路径是什么,同时确保每个切片之间存在重叠.例如:

预期结果

[1,2]
[2,3]
[3,4]
[4,5]
[5,6]
[6,7]
[7,8]
[8,9]
[9,10]
[10,nil]

Answer 1

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
(a + [nil]).each_cons(2).to_a
# => [[1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7],[7, 8], [8, 9], [9, 10], [10, nil]]