optimizing the array stack to shrink array by factor of 1.5, now array stack is faster than linked list stack

pull/1/head
Emir Pasic 9 years ago
parent 50875b13a8
commit 09f75253fe

@ -30,7 +30,7 @@ Collections and data structures found in other languages: Java Collections, C++
**Memory efficient algorithms**:
- Avoiding to keep consume memory by using optimal algorithms and data structures for the given set of problems.
- Avoiding to consume memory by using optimal algorithms and data structures for the given set of problems, e.g. red-black tree in case of TreeMap to avoid keeping redundant sorted array of keys in memory.
**Easy to use library**:

@ -17,7 +17,6 @@ with this distribution for more information.
*/
// Implementation of stack using a slice.
// Use LinkedListStack rather than this Arraystack, because LinkedListStack is a lot faster more memory efficient.
// Structure is not thread safe.
// References: http://en.wikipedia.org/wiki/Stack_%28abstract_data_type%29
@ -35,26 +34,37 @@ func assertInterfaceImplementation() {
type Stack struct {
items []interface{}
top int
}
// Instantiates a new empty stack
func New() *Stack {
return &Stack{}
return &Stack{top: -1}
}
// Pushes a value onto the top of the stack
func (stack *Stack) Push(value interface{}) {
stack.items = append(stack.items, value)
// Increase when capacity is reached by a factor of 1.5 and add one so it grows when size is zero
if stack.top+1 >= cap(stack.items) {
currentSize := len(stack.items)
sizeIncrease := int(1.5*float32(currentSize) + 1.0)
newSize := currentSize + sizeIncrease
newItems := make([]interface{}, newSize, newSize)
copy(newItems, stack.items)
stack.items = newItems
}
stack.top += 1
stack.items[stack.top] = value
}
// Pops (removes) top element on stack and returns it, or nil if stack is empty.
// Second return parameter is true, unless the stack was empty and there was nothing to pop.
func (stack *Stack) Pop() (value interface{}, ok bool) {
size := len(stack.items)
if size > 0 {
value = stack.items[size-1]
stack.items = append([]interface{}(nil), stack.items[:size-1]...)
return value, true
if stack.top >= 0 {
value, ok = stack.items[stack.top], true
// TODO shrink slice at some point
stack.top -= 1
return
}
return nil, false
}
@ -62,21 +72,20 @@ func (stack *Stack) Pop() (value interface{}, ok bool) {
// Returns top element on the stack without removing it, or nil if stack is empty.
// Second return parameter is true, unless the stack was empty and there was nothing to peek.
func (stack *Stack) Peek() (value interface{}, ok bool) {
size := len(stack.items)
if size > 0 {
return stack.items[size-1], true
if stack.top >= 0 {
return stack.items[stack.top], true
}
return nil, false
}
// Returns true if stack does not contain any elements.
func (stack *Stack) Empty() bool {
return len(stack.items) == 0
return stack.Size() == 0
}
// Returns number of elements within the stack.
func (stack *Stack) Size() int {
return len(stack.items)
return stack.top + 1
}
func (stack *Stack) String() string {

@ -72,8 +72,8 @@ func TestArrayStack(t *testing.T) {
}
func BenchmarkArrayStack(b *testing.B) {
// Slow in comparison to the LinkedListStack
// BenchmarkArrayStack 50 31760994 ns/op 8540863 B/op 2010 allocs/op
// Faster in comparison to the LinkedListStack
// BenchmarkArrayStack 5000 325010 ns/op 71648 B/op 1009 allocs/op
// BenchmarkLinkedListStack 5000 390812 ns/op 40016 B/op 2001 allocs/op
for i := 0; i < b.N; i++ {
stack := New()

@ -72,8 +72,8 @@ func TestLinkedListStack(t *testing.T) {
}
func BenchmarkLinkedListStack(b *testing.B) {
// Faster in comparison to the ArrayStack
// BenchmarkArrayStack 50 31760994 ns/op 8540863 B/op 2010 allocs/op
// Slower in comparison to the ArrayStack
// BenchmarkArrayStack 5000 325010 ns/op 71648 B/op 1009 allocs/op
// BenchmarkLinkedListStack 5000 390812 ns/op 40016 B/op 2001 allocs/op
for i := 0; i < b.N; i++ {
stack := New()

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