我实现了两个函数，用于将输入数组向右旋转n次.
如果在所有旋转之后，初始数组将等于生成的数组，则两种实现都会提前退出，而不执行任何操作.每当n为0或数组长度的倍数时，就会发生这种情况.
func rotateRight1(nums []int, n int) {
n = n % len(nums)
if n == 0 {
return
}
lastNDigits := make([]int, n)
copy(lastNDigits, nums[len(nums)-n:])
copy(nums[n:], nums[:len(nums)-n])
copy(nums[:n], lastNDigits)
}
和
func rotateRight2(nums []int, n int) {
n = n % len(nums)
if n == 0 {
return
}
i := 0
current := nums[i]
iAlreadySeen := i
for j := 0; j < len(nums); j++ {
nextI := (i + n) % len(nums)
nums[nextI], current = current, nums[nextI]
i = nextI
// h和le even length arrays where i+k might equal an already seen index
if nextI == iAlreadySeen {
i = (i + 1) % len(nums)
iAlreadySeen = i
current = nums[i]
}
}
}
在进行基准测试时，我惊讶地发现，当两个函数的n等于0时，速度相差+20倍.
func BenchmarkRotateRight1(b *testing.B) {
nums := make([]int, 5_000)
b.ResetTimer()
b.ReportAllocs()
for i := 0; i < b.N; i++ {
rotateRight1(nums, 0)
}
}
func BenchmarkRotateRight2(b *testing.B) {
nums := make([]int, 5_000)
b.ResetTimer()
b.ReportAllocs()
for i := 0; i < b.N; i++ {
rotateRight2(nums, 0)
}
}
go test -bench=.始终会产生这样的结果:
cpu: Intel(R) Core(TM) i7-7500U CPU @ 2.70GHz
BenchmarkRotateRight1-4 1000000000 0.4603 ns/op 0 B/op 0 allocs/op
BenchmarkRotateRight2-4 97236492 12.11 ns/op 0 B/op 0 allocs/op
PASS
I don't underst和 this performance difference as both functions are basically doing the same thing 和 exiting early in the if k == 0 condition.
Could someone help me underst和 this?
go version go1.18 linux/amd64