题目:
Design a max stack that supports push, pop, top, peekMax and popMax.
- push(x) -- Push element x onto stack.
- pop() -- Remove the element on top of the stack and return it.
- top() -- Get the element on the top.
- peekMax() -- Retrieve the maximum element in the stack.
- 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:
- -1e7 <= x <= 1e7
- Number of operations won't exceed 10000.
- The last four operations won't be called when stack is empty.
思路:
我们定义两个数据结构:1)list<int>用来模拟栈,这样可以实现push,pop和top;2)map<int, vector<list<int>::iterator>>,用来实现peekMax和popMax,因为我们可以从map中利用O(1)的时间复杂度得到最大的数,而由于我们在map中也同时保存了各个数对应的迭代器,所以也可以在O(1)的时间复杂度内删除掉list中的内容。
这样的话,可以保证push和pop的时间复杂度为O(logn),top,peekMax,popMax的时间复杂度都为O(1)。整个算法的空间复杂度为O(n)。我感觉应该是最优的算法了。
代码:
class MaxStack {
public:
/** initialize your data structure here. */
MaxStack() {
}
void push(int x) {
auto it = l.insert(l.end(), x);
m[*it].push_back(it);
}
int pop() {
auto it = l.end();
--it;
int value = *it;
m[*it].pop_back();
if (m[*it].empty()) {
m.erase((*it));
}
l.erase(it);
return value;
}
int top() {
return *(l.rbegin());
}
int peekMax() {
return m.rbegin()->first;
}
int popMax() {
auto it = m.rbegin()->second.back();
int value = m.rbegin()->first;
m.rbegin()->second.pop_back();
if (m.rbegin()->second.empty()) {
m.erase(m.rbegin()->first);
}
l.erase(it);
return value;
}
private:
list<int> l;
map<int, vector<list<int>::iterator>> m;
};
/**
* 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();
*/