实现返回栈中的最小元素 要求 时间 O(1) 

    1. package main
    3. import "fmt"
    5. type Stack struct {
    6.     stack []item
    7. }
    9. type item struct {
    10.     min, x int
    11. }
    13. func NewStack() Stack {
    14.     return Stack{}
    15. }
    17. func(m *Stack) Push(x int) {
    18.     min := x
    19.     if len(m.stack) > 0 && m.GetMin() < x {
    20.         min = m.GetMin()
    21.     }
    22.     m.stack = append(m.stack, item{min: min, x: x})
    23.     return
    24. }
    26. func (m *Stack) Pop() {
    27.     length := len(m.stack)
    28.     m.stack = m.stack[:length - 1]
    29.     return
    30. }
    32. func (m *Stack) Top() int {
    33.     length := len(m.stack)
    34.     return m.stack[length - 1].x
    35. }
    37. func (m *Stack) GetMin() int {
    38.     length := len(m.stack)
    39.     return m.stack[length - 1].min
    40. }
    42. func main() {
    43.     minStack := new(Stack)
    44.     minStack.Push(-2)
    45.     minStack.Push(0)
    46.     minStack.Push(3)
    47.     minStack.Push(-5)
    49.     fmt.Println(minStack.GetMin())
    50.     minStack.Pop()
    52.     fmt.Println(minStack.Top())
    53.     fmt.Println(minStack.GetMin())
    54. }