英文:
Is there a Count function in Go but for overlapping substrings?
问题
我正在使用Go语言的strings.Count(s, subtring)函数，它工作得很好，但它只能找到字符串s中的非重叠的子字符串实例。是否有一个函数可以找到重叠的实例呢？
例如：
s := "ccc"
fmt.Println(strings.Count(s, "cc")) // 返回1，但我想要2（有两个子字符串s[0:1]和s[1:2]）
谢谢。
英文:
Im using Go's strings.Count(s, subtring) function which works fine but it finds only the non-overlapping instances of subtring in s. Is there a function which finds the overlapping instances as well?
So for example
s := "ccc"
fmt.Println(strings.Count(s, "cc")) // returns 1 but I want 2 (there are 2 substrings s[0:1] and s[1:2])
Thanks
答案1
得分: 3
这似乎可以实现：
package main
import "strings"
func count(sub, pat string) int {
var c int
for d := range sub {
if strings.HasPrefix(sub[d:], pat) { c++ }
}
return c
}
func main() {
c := count("ccc", "cc")
println(c == 2)
}
请注意，如果Go语言有正则表达式的前瞻功能，你可以像Python一样做：
>>> import re
>>> len(re.findall('(?=cc)', 'ccc'))
2
英文:
This seems to do it:
package main
import "strings"
func count(sub, pat string) int {
var c int
for d := range sub {
if strings.HasPrefix(sub[d:], pat) { c++ }
}
return c
}
func main() {
c := count("ccc", "cc")
println(c == 2)
}
Note if Go had Regular Expression Lookahead, you could do something like this
Python:
>>> import re
>>> len(re.findall('(?=cc)', 'ccc'))
2
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