Hi community,
I improved my binary search example with both sort.Search and sort.Find.
Please correct me if I got new wrongs with new example.
1.
My original question was if there are duplicate entries in sorted slice.
After I located the first dup, how can I go to the rest of them without loop.
I ever thought using loop was ugly, so I tried to avoid loop.
But I think loop is ok. because dups are adjacent on sorted slice.
2.
How can I use tuple style comparison found in other languages, like:
(a.name, a.num) > (b.name, b.num)
struct can be used for == only, not for <, >
3.
I think my conditions covered all the cases, but I need an extra trailing
return ?
Whats the right way to do it?
if a < b {
return -1
} else if a == b {
return 0
} else if a > b {
return 1
}
return -999 // an extra return
Thanks
//---
//sort.Search example
func sortSearch() {
haystack := []T{
{"aaa", 111},
{"bbb", 222}, //dup
{"bbb", 222}, //dup
{"ccc", 333},
}
needle := T{"bbb", 222}
index := sort.Search(len(haystack), func(i int) bool {
//asc >=
//desc <=
if haystack[i].name >= needle.name ||
haystack[i].name == needle.name &&
haystack[i].num >= needle.num {
return true;
}
return false
})
log.Println(index)
for i := index; i < len(haystack); i++ {
if haystack[i] == needle {
log.Println(i, haystack[i])
} else {
break
}
}
}
//sort.Find example
func sortFind() {
log.SetFlags(log.LstdFlags | log.Llongfile)
haystack := []T{
{"aaa", 111},
{"bbb", 222}, //dup
{"bbb", 222}, //dup
{"ccc", 333},
}
needle := T{"bbb", 222}
i, found := sort.Find(len(haystack), func(i int) int {
//asc -1 0 1
//desc 1 0 -1
if needle.name < haystack[i].name ||
needle.name == haystack[i].name &&
needle.num < haystack[i].num {
return -1
} else
if needle == haystack[i] {
return 0
} else
if needle.name > haystack[i].name ||
needle.name == haystack[i].name &&
needle.num > haystack[i].num {
return 1
}
return 9 // do i have to add this line // Line 66
})
log.Println(found, i)
for i := i; found && i != len(haystack); i++ {
if haystack[i] == needle {
log.Println(i, haystack[i])
} else {
break
}
}
}
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