LeetCode #716 - Max Stack

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本文介绍了一种名为MaxStack的数据结构设计，支持push、pop、top、peekMax和popMax操作。通过双栈实现，其中一个栈保持所有元素的压栈顺序，另一个栈保持栈顶为最大元素，确保了在O(1)时间内完成peekMax和popMax操作。

题目描述：

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.

Example 1:

MaxStack stack = new MaxStack();

stack.push(5);

stack.push(1);

stack.push(5);

stack.top(); -> 5

stack.popMax(); -> 5

stack.top(); -> 1

stack.peekMax(); -> 5

stack.pop(); -> 1

stack.top(); -> 5

Note:

1. -1e7 <= x <= 1e7

2. Number of operations won't exceed 10000.

3. The last four operations won't be called when stack is empty.

class MaxStack {
public:
    /** initialize your data structure here. */
    MaxStack() {}
    
    void push(int x) {
        if(s1.empty()) s1.push(x);
        else s1.push(max(s1.top(),x));
        s2.push(x);
    }
    
    int pop() {
        int x=s2.top();
        s1.pop();
        s2.pop();
        return x;
    }
    
    int top() {
        return s2.top();
    }
    
    int peekMax() {
        return s1.top();
    }
    
    int popMax() {
        int x=s1.top();
        stack<int> temp;
        while(!s2.empty()&&s1.top()!=s2.top())
        {
            temp.push(s2.top());
            s1.pop();
            s2.pop();
        }
        s1.pop();
        s2.pop();
        while(!temp.empty()) 
        {
            push(temp.top());
            temp.pop();
        }
        return x;
    }
    
private:
    stack<int> s1; //保持栈顶为最大元素
    stack<int> s2; //按照压栈顺序保持所有元素
};