Golang实现最大堆/最小堆

参考： https://yangjiahao106.github.io/2019/01/15/golang-%E6%9C%80%E5%A4%A7%E5%A0%86%E5%92%8C%E6%9C%80%E5%B0%8F%E5%A0%86/
https://studygolang.com/articles/24288

方法一

此方法就是比较传统、常见的方法，下面来构建一个最小堆：

type Heap struct {Size intElems []int
}func NewHead(maxSize int) *Heap {//minHead := new(Heap)//minHead.Elems = make([]int, maxSize, maxSize)//return minHeadminHead := Heap{Size: -1, Elems: make([]int, maxSize, maxSize)}return &minHead
}func (h *Heap) Push(x int)  {if h.Size == len(h.Elems) - 1 {return}h.Size++i := h.Size// 与父节点进行比较，如果小于父节点，则向上冒泡；否则，停止；for i > 0 {parent := (i - 1) / 2if h.Elems[parent] <= x {break}h.Elems[i] = h.Elems[parent]i = parent}h.Elems[i] = x
}func (h *Heap) Pop() int {if h.Size < 0 {return math.MinInt32}ret := h.Elems[0]// 将最后一个节点提到根节点，然后向下交换x := h.Elems[h.Size]i := 0for {l := 2*i + 1r := 2*i + 2if l >= h.Size {break}if r < h.Size && h.Elems[l] > h.Elems[r] {l = r}if x <= h.Elems[l] {break}h.Elems[i] = h.Elems[l]i = l}h.Size--h.Elems[i] = xreturn ret
}

测试：

func main() {minHead := NewHead(10)minHead.Push(2)minHead.Push(7)minHead.Push(5)minHead.Push(5)minHead.Push(4)minHead.Push(8)minHead.Push(8)minHead.Push(8)minHead.Push(8)minHead.Push(12)minHead.Push(20)k := minHead.Pop()for k != math.MinInt32 {fmt.Println(k)k = minHead.Pop()}
}/* output:24557888812
*/

一些其他的操作，可以在此基础进行扩充。

方法二

方法二看似要比方法一简单，首先需要介绍一下container/heap包，heap包对任意实现heap接口的类型提供堆操作。heap包中包含如下接口：

type Interface interface {sort.InterfacePush(x interface{}) // add x as element Len()Pop() interface{}   // remove and return element Len() - 1.
}// sort.Interface
type Interface interface {Len() intLess(i, j int) boolSwap(i, j int)
}

这里需要解释的应该只有Less(i, j int) bool 接口，我们查看sort.go源码时，会看到很多类似于下面这种代码：

if data.Less(m1, m0) {data.Swap(m1, m0)
}

所以，结果很明显，代码会根据Less()的结果来决定是否要交换data[i]和data[j]。
接下来，我们实现对应的接口：

type minHeap []intfunc (h minHeap) Len() int {return len(h)
}// 这里实现了小根堆，如果想要实现大根堆可以改为 h[i] > h[j]
func (h minHeap) Less(i, j int) bool {return h[i] < h[j]
}func (h *minHeap) Swap(i, j int)  {(*h)[i], (*h)[j] = (*h)[j], (*h)[i]
}func (h *minHeap) Push(x interface{})  {*h = append(*h, x.(int))
}func (h *minHeap) Pop() interface{} {res := (*h)[len(*h) - 1]*h = (*h)[:len(*h) - 1]return res
}

测试：

func main() {h := make(minHeap, 0)heap.Init(&h)heap.Push(&h, 2)heap.Push(&h, 7)heap.Push(&h, 5)heap.Push(&h, 5)heap.Push(&h, 4)heap.Push(&h, 6)for len(h) != 0 {fmt.Println(heap.Pop(&h))}
}/*output:245567
*/