itest: even out num of tests per tranche

Previous splitting logic simply put all the remainder in the last
tranche, which could make the last tranche run significantly more test
cases. We now change it so the remainder is evened out across tranches.

itest/lnd_test.go

@@ -174,6 +174,30 @@ func maybeShuffleTestCases() {
 	})
 }
 
+// createIndices divides the number of test cases into pairs of indices that
+// specify the start and end of a tranche.
+func createIndices(numCases, numTranches uint) [][2]uint {
+	// Calculate base value and remainder.
+	base := numCases / numTranches
+	remainder := numCases % numTranches
+
+	// Generate indices.
+	indices := make([][2]uint, numTranches)
+	start := uint(0)
+
+	for i := uint(0); i < numTranches; i++ {
+		end := start + base
+		if i < remainder {
+			// Add one for the remainder.
+			end++
+		}
+		indices[i] = [2]uint{start, end}
+		start = end
+	}
+
+	return indices
+}
+
 // getTestCaseSplitTranche returns the sub slice of the test cases that should
 // be run as the current split tranche as well as the index and slice offset of
 // the tranche.
@@ -200,12 +224,9 @@ func getTestCaseSplitTranche() ([]*lntest.TestCase, uint, uint) {
 	maybeShuffleTestCases()
 
 	numCases := uint(len(allTestCases))
-	testsPerTranche := numCases / numTranches
-	trancheOffset := runTranche * testsPerTranche
-	trancheEnd := trancheOffset + testsPerTranche
-	if trancheEnd > numCases || runTranche == numTranches-1 {
-		trancheEnd = numCases
-	}
+	indices := createIndices(numCases, numTranches)
+	index := indices[runTranche]
+	trancheOffset, trancheEnd := index[0], index[1]
 
 	return allTestCases[trancheOffset:trancheEnd], threadID,
 		trancheOffset