剑指 Offer 59 - II. 队列的最大值
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 的均摊时间复杂度都是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
type MaxQueue struct {
queue []int
max []int
}
func Constructor() MaxQueue {
return MaxQueue{
queue : make([]int, 0),
max: make([]int, 0),
}
}
func (this *MaxQueue) Max_value() int {
if len(this.queue) == 0 {return -1}
return this.max[0]
}
func (this *MaxQueue) Push_back(value int) {
this.queue = append(this.queue, value)
for len(this.max) > 0 && value > this.max[len(this.max)-1] {
this.max = this.max[:len(this.max)-1]
}
this.max = append(this.max, value)
}
func (this *MaxQueue) Pop_front() int {
if len(this.queue) == 0 {return -1}
tmp := this.queue[0]
if this.max[0] == tmp {
this.max = this.max[1:]
}
this.queue = this.queue[1:]
return tmp
}
/**
* 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();
*/