ElixirでBinary Searchを実装してみた。メモ。
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| defmodule Search do | |
| def binary(_, []), do: -1 | |
| def binary(search, [head|tail]) when tail == [], do: if search == head, do: 0, else: -1 | |
| def binary(search, list), do: binary search, 0, length(list) – 1, list | |
| defp binary(search, left_len, right_len, list) when right_len – left_len == 1 do | |
| cond do | |
| Enum.at(list, left_len) == search -> | |
| left_len | |
| Enum.at(list, right_len) == search -> | |
| right_len | |
| true -> | |
| -1 | |
| end | |
| end | |
| defp binary(search, left_len, right_len, list) do | |
| mid = round((right_len – left_len) / 2) | |
| cond do | |
| Enum.at(list, mid) > search -> | |
| binary search, left_len, mid, list | |
| true -> | |
| binary search, mid, right_len, list | |
| end | |
| end | |
| end | |
| ExUnit.start | |
| defmodule SearchTest do | |
| use ExUnit.Case | |
| test "empty case" do | |
| assert Search.binary(1, []) == -1 | |
| end | |
| test "one case" do | |
| assert Search.binary(1, [1]) == 0 | |
| assert Search.binary(2, [1]) == -1 | |
| end | |
| test "two case" do | |
| assert Search.binary(1, [1, 2]) == 0 | |
| assert Search.binary(2, [1, 2]) == 1 | |
| assert Search.binary(3, [1, 2]) == -1 | |
| end | |
| test "many case" do | |
| assert Search.binary(1, [1, 2, 5]) == 0 | |
| assert Search.binary(5, [1, 2, 5]) == 2 | |
| assert Search.binary(3, [1, 2, 5]) == -1 | |
| end | |
| end |
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