Hard
讲真这道题说不上Hard吧，对Stack, PriorityQueue稍微熟悉一点就能做,注意几点：
PriorityQueue的constructor，我一开始纠结于那个initial capacity要怎么填，那个comparator需不需要重写，然而还是太不熟悉。initial capacity写一个比较大的数字就可以了，integer默认是从小到大排列的，从大到小排列的话写一个
Collections.reverseOrder()就可以了。关于
popMax()方法，一开始我是没想到怎么做的。心想一个Stack要怎么pop出来里面最大的元素呢？原来是新建一个temp stack, 我们把原stack的peek元素一一pop出来放到temp里面，直到遇到真正需要取的那个max把它pop出去，再把原来暂存到temp stack里的元素重新push回到original stack, 很巧妙的方法.
class MaxStack {
Stack<Integer> stack;
PriorityQueue<Integer> pq;
/** initialize your data structure here. */
public MaxStack() {
stack = new Stack<>();
pq = new PriorityQueue<>(10000, Collections.reverseOrder());
}
public void push(int x) {
stack.push(x);
pq.offer(x);
}
public int pop() {
int top = stack.pop();
pq.remove(top);
return top;
}
public int top() {
return stack.peek();
}
public int peekMax() {
return pq.peek();
}
public int popMax() {
int max = pq.poll();
Stack<Integer> temp = new Stack<>();
while (stack.peek() != max){
temp.push(stack.pop());
}
stack.pop();
while (!temp.isEmpty()){
stack.push(temp.pop());
}
return max;
}
}
/**
* Your MaxStack object will be instantiated and called as such:
* MaxStack obj = new MaxStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.peekMax();
* int param_5 = obj.popMax();
*/
