面试题 03.02. Min Stack LCCI

Problem

How would you design a stack which, in addition to push and pop, has a function min which returns the minimum element? Push, pop and min should all operate in 0(1) time.

Example

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

Solution

使用两个栈，s1记录数据，s2记录当前数据对应的最小值。

当有新元素入栈时，直接将其压入s1；
对于s2，将新元素与其栈顶元素相比较，将二者中较小的压入。

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {

    }
    
    void push(int x) {
        val_s.push(x);
        if(min_s.empty())
        {
            min_s.push(x);
        }
        else
        {
            int curMin = min_s.top();
            if(x < curMin)
                min_s.push(x);
            else
                min_s.push(curMin);
        }

    }
    
    void pop() {
        if(!val_s.empty())
        {
            val_s.pop();
            min_s.pop();
        }
    }
    
    int top() {
        return val_s.top();
    }
    
    int getMin() {
        return min_s.top();
    }

    stack<int> min_s;
    stack<int> val_s;
};

/**
 * 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();
 */

Related problem

面试题59 - II. 队列的最大值