716.Max Stack

Tags: Easy Stack Design

Links: https://www.lintcode.com/problem/max-stack/description（LeetCode上锁，用LintCode代替）

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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.

---

这道题在LintCode竟然是Hard，其实只是在Min Stack的基础上增加了一个要求删除最大值的操作，比如依次推入5，3，3，最大值为5，需要删除5，那么其实可以用一个临时栈将3，3存储，然后删掉5后，再把临时存储的数据加入到原来的栈。另外，这里的pop还需要返回被删除的元素，不再是void了。

class MaxStack {
    stack<int> s1, s2;
public:
    /** initialize your data structure here. */
    MaxStack() {}

    void push(int x) {
        if (s2.empty() || x >= s2.top()) s2.push(x);
        s1.push(x);
    }

    int pop() {
        if (s1.empty()) return -1;

        int res = s1.top();
        s1.pop();
        if (res == s2.top()) s2.pop();
        return res;
    }

    int top() {
        if (s1.empty()) return -1;
        return s1.top();
    }

    int peekMax() {
        if (s2.empty()) return -1;
        return s2.top();
    }

    int popMax() {
        if (s2.empty()) return -1;

        int res = s2.top();
        s2.pop();
        stack<int> tmp;
        while (s1.top() != res) {
            tmp.push(s1.top());
            s1.pop();
        }
        s1.pop();
        while (!tmp.empty()) {
            push(tmp.top());
            tmp.pop();
        }
        return res;
    }
};

/**
 * 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();
 */

数据更新后出错的测试用例：

["MaxStack","push","push","popMax","peekMax"]
[[],[5],[1],[],[]]
Output: [null,null,null,5,-1]
Judge: [null,null,null,5,1]

注意删除最大值需要重新来一次push。