strings: use Index in Count

This simplifies code and provides performance iprovments:
Similar to https://go-review.googlesource.com/#/c/28577

CountHard1-48               1.74ms ±14%  0.17ms ±14%  -90.16%  (p=0.000 n=19+19)
CountHard2-48               1.78ms ±15%  0.25ms ±13%  -86.10%  (p=0.000 n=19+20)
CountHard3-48               1.78ms ±12%  0.80ms ±11%  -55.19%  (p=0.000 n=17+20)
CountTorture-48             13.5µs ±14%  13.6µs ±11%     ~     (p=0.625 n=18+19)
CountTortureOverlapping-48  6.92ms ±13%  8.42ms ±11%  +21.72%  (p=0.000 n=19+17)

Change-Id: Ief120aee918a66487c76be56e0796871c8502f89
Reviewed-on: https://go-review.googlesource.com/28586
Run-TryBot: Ilya Tocar <ilya.tocar@intel.com>
TryBot-Result: Gobot Gobot <gobot@golang.org>
Reviewed-by: Brad Fitzpatrick <bradfitz@golang.org>

src/strings/strings.go

@@ -77,48 +77,18 @@ func hashStrRev(sep string) (uint32, uint32) {
 func Count(s, sep string) int {
 	n := 0
 	// special cases
-	switch {
-	case len(sep) == 0:
+	if len(sep) == 0 {
 		return utf8.RuneCountInString(s) + 1
-	case len(sep) == 1:
-		// special case worth making fast
-		c := sep[0]
-		for i := 0; i < len(s); i++ {
-			if s[i] == c {
-				n++
-			}
 	}
+	offset := 0
+	for {
+		i := Index(s[offset:], sep)
+		if i == -1 {
 			return n
-	case len(sep) > len(s):
-		return 0
-	case len(sep) == len(s):
-		if sep == s {
-			return 1
-		}
-		return 0
 		}
-	// Rabin-Karp search
-	hashsep, pow := hashStr(sep)
-	h := uint32(0)
-	for i := 0; i < len(sep); i++ {
-		h = h*primeRK + uint32(s[i])
-	}
-	lastmatch := 0
-	if h == hashsep && s[:len(sep)] == sep {
-		n++
-		lastmatch = len(sep)
-	}
-	for i := len(sep); i < len(s); {
-		h *= primeRK
-		h += uint32(s[i])
-		h -= pow * uint32(s[i-len(sep)])
-		i++
-		if h == hashsep && lastmatch <= i-len(sep) && s[i-len(sep):i] == sep {
 		n++
-			lastmatch = i
+		offset += i + len(sep)
 	}
-	}
-	return n
 }
 
 // Contains reports whether substr is within s.