本来准备先看堆的,发现引用了sort。等等还是先看ring,再看sort吧。
先看数据结构:
type Ring struct {
next, prev *Ring
Value interface{} // for use by client; untouched by this library
}
简单,跟链表很像。环。
初始化简单:
func (r *Ring) init() *Ring {
r.next = r
r.prev = r
return r
}
迭代函数也比较简单:
func (r *Ring) Next() *Ring {
if r.next == nil {
return r.init()
}
return r.next
}
func (r *Ring) Prev() *Ring {
if r.next == nil {
return r.init()
}
return r.prev
}
move移动函数,小于0往前,大于0往后:
func (r *Ring) Move(n int) *Ring {
if r.next == nil {
return r.init()
}
switch {
case n < 0:
for ; n < 0; n++ {
r = r.prev
}
case n > 0:
for ; n > 0; n-- {
r = r.next
}
}
return n
}
创建n个元素:
func New(n int) *Ring {
if n <= 0 {
return nil
}
r := new(Ring)
p := r
for i := 1; i < n; i++ {
p.next = &Ring{prev: p}
p = p.next
}
p.next = r
r.prev = p
return r
}
link和unlink:
func (r *Ring) Link(s *Ring) *Ring {
n := r.Next()
if s != nil {
p := s.Prev()
// Note: Cannot use multiple assignment because
// evaluation order of LHS is not specified.
r.next = s
s.prev = r
n.prev = p
p.next = n
}
return n
}
func (r *Ring) Unlink(n int) *Ring {
if n <= 0 {
return nil
}
return r.Link(r.Move(n + 1))
}
注意unlink使用了link。
计算环的长度:
func (r *Ring) Len() int {
n := 0
if r != nil {
n = 1
for p := r.Next(); p != r; p = p.next {
n++
}
}
return n
}
时间复杂度为n
稍微特殊一点的就是这个do函数了,对于环中每个元素调用指定func:
func (r *Ring) Do(f func(interface{})) {
if r != nil {
f(r.Value)
for p := r.Next(); p != r; p = p.next {
f(p.Value)
}
}
}