[#697] governance: Make best effort traversing main chain list of keys

We should go through every key in main chain list to merget lists
as fast as possible. Previously we drop main chain traversing as
soon as we have no more new keys to add. Instead we should try
to go for old keys in the list and add it as more as we can.

Signed-off-by: Alex Vanin <alexey@nspcc.ru>

pkg/innerring/processors/governance/list.go

@@ -44,19 +44,24 @@ func newAlphabetList(sidechain, mainnet keys.PublicKeys) (keys.PublicKeys, error
 	newNodes := 0
 	newNodeLimit := (ln - 1) / 3
 
-	for i := 0; i < ln; i++ {
-		if newNodes == newNodeLimit {
+	for _, node := range mainnet {
+		if len(result) == ln {
 			break
 		}
 
-		mainnetAddr := mainnet[i].Address()
+		limitReached := newNodes == newNodeLimit
+
+		mainnetAddr := node.Address()
 		if _, ok := hmap[mainnetAddr]; !ok {
+			if limitReached {
+				continue
+			}
 			newNodes++
 		} else {
 			hmap[mainnetAddr] = true
 		}
 
-		result = append(result, mainnet[i])
+		result = append(result, node)
 	}
 
 	if newNodes == 0 {

pkg/innerring/processors/governance/list_test.go

@@ -68,6 +68,15 @@ func TestNewAlphabetList(t *testing.T) {
 		require.NoError(t, err)
 		require.True(t, equalPublicKeyLists(exp, got)) // expect {1, 2, 4, 5}, not {1, 2, 3, 5}
 	})
+
+	t.Run("new keys in the middle", func(t *testing.T) {
+		orig := keys.PublicKeys{k[0], k[1], k[2], k[6], k[7], k[8], k[9]}
+		// `exp` should contain maximum amount of new keys (2) in the middle
+		exp := keys.PublicKeys{k[0], k[3], k[4], k[6], k[7], k[8], k[9]}
+		got, err := newAlphabetList(orig, exp)
+		require.NoError(t, err)
+		require.True(t, equalPublicKeyLists(exp, got))
+	})
 }
 
 func TestUpdateInnerRing(t *testing.T) {