# 716. Max Stack (E)

Description

Design a max stack that supports push, pop, top, peekMax and popMax.

1. push(x) -- Push element x onto stack.
2. pop() -- Remove the element on top of the stack and return it.
3. top() -- Get the element on the top.
4. peekMax() -- Retrieve the maximum element in the stack.
5. popMax() -- Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.
6. `-1e7 <= x <= 1e7`
7. Number of operations won't exceed `10000`.
8. The last four operations won't be called when stack is empty.

Example

```
Input:
push(5)
push(1)
push(5)
top()
popMax()
top()
peekMax()
pop()
top()
Output:
5
5
1
5
1
5
```

### Solution:

<https://aaronice.gitbook.io/lintcode/stack/max-stack>\
使用两个栈来模拟，s1为普通的栈，用来保存所有的数字，而s2为最大栈，用来保存出现的最大的数字。

在push()函数中，我们先来看s2，如果s2为空，或者s2的栈顶元素小于等于x，将x压入s2中。因为s2保存的是目前为止最大的数字，所以一旦新数字大于等于栈顶元素，说明遇到更大的数字了，压入栈。然后将数组压入s1，s1保存所有的数字，所以都得压入栈。

在pop()函数中，当s2的栈顶元素和s1的栈顶元素相同时，我们要移除s2的栈顶元素，因为一个数字不在s1中了，就不能在s2中。然后取出s1的栈顶元素，并移除s1，返回即可。

在top()函数中，直接返回s1的top()函数即可。

在peekMax()函数中，直接返回s2的top()函数即可。

在popMax()函数中，先将s2的栈顶元素保存到一个变量mx中，然后我们要在s1中删除这个元素，由于栈无法直接定位元素，所以我们用一个临时栈t，将s1的出栈元素保存到临时栈t中，当s1的栈顶元素和s2的栈顶元素相同时退出while循环，此时我们在s1中找到了s2的栈顶元素，分别将s1和s2的栈顶元素移除，然后要做的是将临时栈t中的元素加回s1中，**注意此时容易犯的一个错误是，没有同时更新s2**，所以我们直接调用push(), pop()函数即可

```
class MaxStack {
    Stack<Integer> stack;
    Stack<Integer> maxStack;

    public MaxStack() {
        stack = new Stack();
        maxStack = new Stack();
    }

    public void push(int x) {
        int max = maxStack.isEmpty() ? x : maxStack.peek();
        maxStack.push(max > x ? max : x);
        stack.push(x);
    }

    public int pop() {
        maxStack.pop();
        return stack.pop();
    }

    public int top() {
        return stack.peek();
    }

    public int peekMax() {
        return maxStack.peek();
    }

    public int popMax() {
        int max = peekMax();
        Stack<Integer> buffer = new Stack();
        // 千万注意下面要直接调用push(),pop() 以便同时更新两个栈
        while (top() != max) buffer.push(pop());
        pop();
        while (!buffer.isEmpty()) push(buffer.pop());
        return max;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://junnie.gitbook.io/nine-chapter/8.data-structure/716.-max-stack-e.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.