请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的时间复杂度都是O(1)。
若队列为空，pop_front 和 max_value 需要返回 -1
示例 1：
输入:
["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
[[],[1],[2],[],[],[]]
输出: [null,null,null,2,1,2]
示例 2：
输入:
["MaxQueue","pop_front","max_value"]
[[],[],[]]
输出: [null,-1,-1]
限制：
1 <= push_back,pop_front,max_value的总操作数 <= 10000
1 <= value <= 10^5
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/dui-lie-de-zui-da-zhi-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
队列的基础知识
type MaxQueue struct {
head int
tail int
count int
max int
list []int
}
func Constructor() MaxQueue {
q := MaxQueue{
head : 0,
tail : 0,
count :0,
max : 0,
list : make([]int ,0),
}
return q
}
func (this *MaxQueue) Max_value() int {
if this.count == 0 {
return -1
}
// fmt.Println("调用Max_Value:",this.max)
return this.max
}
func (this *MaxQueue) Push_back(value int) {
//更新max
if this.count == 0 {
this.max = value
} else if value > this.max {
this.max = value
}
this.list = append(this.list,value)
this.tail++
this.count++
// fmt.Println(this.max,this.list)
// fmt.Println("调用Push_Back"," ","value: ",value," ",this.list," ",this.max)
}
func (this *MaxQueue) Pop_front() int {
if this.count == 0{
return -1
}
pop := this.list[0]
this.count--
this.head++
this.list = this.list[this.head:]
this.head = 0
//需要更新max
if pop == this.max && this.count > 0{
tmp := this.list[len(this.list)-1]
for _,v := range this.list {
if v > tmp {
tmp = v
}
}
this.max = tmp
} else if this.count == 0{
this.max = 0
}
// fmt.Println("调用pop"," ",this.list," ",this.max," ",pop)
return pop
}
/**
* Your MaxQueue object will be instantiated and called as such:
* obj := Constructor();
* param_1 := obj.Max_value();
* obj.Push_back(value);
* param_3 := obj.Pop_front();
*/代码中带了调试用的语句，可直接打开注释进行查看输出
