regexp: add Split

As discussed in issue 2672 and on golang-nuts, this CL adds a Split() method
to regexp. It is based on returning the "opposite" of FindAllString() so that
the returned substrings are everything not matched by the expression.

See: https://groups.google.com/forum/?fromgroups=#!topic/golang-nuts/xodBZh9Lh2E

Fixes #2762.

R=remyoudompheng, r, rsc
CC=golang-dev
https://golang.org/cl/6846048

src/pkg/regexp/regexp.go

@@ -1048,3 +1048,52 @@ func (re *Regexp) FindAllStringSubmatchIndex(s string, n int) [][]int {
 	}
 	return result
 }
+
+// Split slices s into substrings separated by the expression and returns a slice of
+// the substrings between those expression matches.
+//
+// The slice returned by this method consists of all the substrings of s
+// not contained in the slice returned by FindAllString. When called on an expression
+// that contains no metacharacters, it is equivalent to strings.SplitN.
+//
+// Example:
+//   s := regexp.MustCompile("a*").Split("abaabaccadaaae", 5)
+//   // s: ["", "b", "b", "c", "cadaaae"]
+//
+// The count determines the number of substrings to return:
+//   n > 0: at most n substrings; the last substring will be the unsplit remainder.
+//   n == 0: the result is nil (zero substrings)
+//   n < 0: all substrings
+func (re *Regexp) Split(s string, n int) []string {
+
+	if n == 0 {
+		return nil
+	}
+
+	if len(re.expr) > 0 && len(s) == 0 {
+		return []string{""}
+	}
+
+	matches := re.FindAllStringIndex(s, n)
+	strings := make([]string, 0, len(matches))
+
+	beg := 0
+	end := 0
+	for _, match := range matches {
+		if n > 0 && len(strings) >= n-1 {
+			break
+		}
+
+		end = match[0]
+		if match[1] != 0 {
+			strings = append(strings, s[beg:end])
+		}
+		beg = match[1]
+	}
+
+	if end != len(s) {
+		strings = append(strings, s[beg:])
+	}
+
+	return strings
+}