# 155.Min Stack

难度:Easy

**155.Min Stack**

> 设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。

push(x) -- 将元素 x 推入栈中。 pop() -- 删除栈顶的元素。 top() -- 获取栈顶元素。 getMin() -- 检索栈中的最小元素。 示例:

MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.getMin(); --> 返回 -2.

要设计一个支持返回最小值的栈，可以使用两个栈实现：一个支持push和pop、top等操作，另一个保存每次的最小值。

```
class MinStack {
private:
        stack<int> num;
        stack<int > m;
public:
    /** initialize your data structure here. */
    MinStack() {
 
        
    }
    
    void push(int x) {
        if(num.empty()) m.push(x);
        else m.push(min(m.top(),x));
        num.push(x);
        
        
    }
    
    void pop() {
        num.pop();
        m.pop();
    }
    
    int top() {
        return num.top();
        
    }
    
    int getMin() {
        return m.top();
    }
};

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack* obj = new MinStack();
 * obj->push(x);
 * obj->pop();
 * int param_3 = obj->top();
 * int param_4 = obj->getMin();
 */
```

> 执行用时 :36 ms, 在所有 C++ 提交中击败了97.28%的用户\
> 内存消耗 :16.9MB, 在所有 C++ 提交中击败了69.68%的用户


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