LeetCode(155) Min Stack

题目如下：

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.

分析如下：

1 想到用两个stack来维护这个结构。1个stack用来正常进行stack的push pop等操作。另外1个stack用来维护min.每次对stack进行pop或者push时，也对min_stack进行相应操作。

2 第2个stack的大小是可以进行优化的。不一定每个min都要入栈min_stack,push的时候，只入栈比当前min小或者相等的值就可以了。pop的时候，比较待pop元素和min_stack的top的大小。如果待pop元素和min_stack top相等，则将min stack进行pop。

3 online judge做得真心好，如果用了两个栈但是没有空间优化，是没法ac的。

直接把官网上的unlocked之后的解答抄过来:

O(1) runtime, O(n) space – Extra stack:

Use an extra stack to keep track of the current minimum value. During the push operation we choose the new element or the current minimum, whichever that is smaller to push onto the min stack.


O(1) runtime, O(n) space – Minor space optimization:

If a new element is larger than the current minimum, we do not need to push it on to the min stack. When we perform the pop operation, check if the popped element is the same as the current minimum. If it is, pop it off the min stack too.

我的代码：

class MinStack {
private:
    std::stack<int> stack;
    std::stack<int> min_stack;
public:
    void push(int x) {
        stack.push(x);
        if (min_stack.empty() || ((!min_stack.empty()) && x <= min_stack.top())) {  //NOTE: 是“小于等于”，不是“小于”
            min_stack.push(x);
        }
    }
    
    void pop() {
        if (!stack.empty()) {
            if (stack.top() == min_stack.top())
                min_stack.pop();
            stack.pop();
        }
    }
    
    int top() {
        if (!stack.empty())
            return stack.top();
    }
    
    int getMin() {
        if (!min_stack.empty())
            return min_stack.top();
    }
};