leetcode 716

问题

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:

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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.

分析

这道题中一开始没有说清楚，还以为和MinStack那道题类似，直接用两个栈解决即可。后来发现popMax函数只会把栈中最大的元素pop出来，而其他元素还都留在栈中。这样的话就不能使用stack等线性表来实现了，改用双向链表list来实现。然后在通过一个 map\::iterator>> 来保存每个值对应的双向链表的地址。map默认为从小到大的排序。因此，在popMax的时候，只需要将map的最后一个元素对应的栈取一个栈顶，并将其在list中移除即可。

代码

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class MaxStack {
public:
    list<int> l;  // simulate stack
    map<int, stack<list<int>::iterator>> mp;
    /** initialize your data structure here. */
    MaxStack() {
        l.clear();
        mp.clear();
    }
    
    void push(int x) {
        l.push_back(x);
        auto it = l.end();
        mp[x].push(--it);
    }
    
    int pop() {
        int x = l.back();
        l.pop_back();
        mp[x].pop();
        if (mp[x].size() == 0)
            mp.erase(x);
        return x;
    }
    
    int top() {
        return l.back();
    }
    
    int peekMax() {
        auto it = mp.end();
        return (--it)->first;
    }
    
    int popMax() {
        auto it = mp.end();
        int x = (--it)->first;
        l.erase(it->second.top());
        it->second.pop();
        if (it->second.size() == 0)
            mp.erase(it);
        return x;
    }
};

/**
 * 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();
 */