Bug Fix Enumerator::Lazy#uniq state for multiple call

* enumerator.c (lazy_uniq_i): create new hash for each calls.
  [Fix GH-1820]

Currently

	2.5.0-preview1 :001 > arr = (0..100).lazy.uniq{|i| i % 10}
	 => #<Enumerator::Lazy: #<Enumerator::Lazy: 0..100>:uniq>
	2.5.0-preview1 :002 > arr.to_a
	 => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
	2.5.0-preview1 :003 > arr.to_a
	 => []

Expected

arr.to_a to always return same output

From: Anmol Chopra <anmolchopra@rocketbox.in>

git-svn-id: svn+ssh://ci.ruby-lang.org/ruby/trunk@62731 b2dd03c8-39d4-4d8f-98ff-823fe69b080e

spec/ruby/core/enumerator/lazy/uniq_spec.rb

@@ -3,6 +3,26 @@ require_relative 'fixtures/classes'
 
 ruby_version_is '2.4' do
   describe 'Enumerator::Lazy#uniq' do
+    context 'without block' do
+      before :each do
+        @lazy = [0, 1, 0, 1].to_enum.lazy.uniq
+      end
+
+      it 'returns a lazy enumerator' do
+        @lazy.should be_an_instance_of(Enumerator::Lazy)
+        @lazy.force.should == [0, 1]
+      end
+
+      it 'return same value after rewind' do
+        @lazy.force.should == [0, 1]
+        @lazy.force.should == [0, 1]
+      end
+
+      it 'sets the size to nil' do
+        @lazy.size.should == nil
+      end
+    end
+
     context 'when yielded with an argument' do
       before :each do
         @lazy = [0, 1, 2, 3].to_enum.lazy.uniq(&:even?)
@@ -13,6 +33,11 @@ ruby_version_is '2.4' do
         @lazy.force.should == [0, 1]
       end
 
+      it 'return same value after rewind' do
+        @lazy.force.should == [0, 1]
+        @lazy.force.should == [0, 1]
+      end
+
       it 'sets the size to nil' do
         @lazy.size.should == nil
       end
@@ -31,6 +56,12 @@ ruby_version_is '2.4' do
         @lazy = enum.lazy
       end
 
+      it 'return same value after rewind' do
+        enum = @lazy.uniq { |_, label| label.downcase }
+        enum.force.should == [[0, 'foo'], [2, 'bar']]
+        enum.force.should == [[0, 'foo'], [2, 'bar']]
+      end
+
       it 'returns all yield arguments as an array' do
         @lazy.uniq { |_, label| label.downcase }.force.should == [[0, 'foo'], [2, 'bar']]
       end