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.

Example:

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

最小棧，要求getMin()為檢索當前stack中的最小值。

每次添加元素時用minValue來維護一個最小值

在彈出元素時，如果彈出的是最小元素，則需要在彈出後的stack中找出最小值，因為stack不能遍歷，所以使用另一個tmpStack來存儲原stack中彈出的每一個值找出最小元素。在將tmpStack中的值再傳給原stack中。權當是一個遍歷。

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {
        minValue = INT_MAX;
    }
    
    
 
void push(int x) {
        s.push(x);
        minValue = min(minValue, x);
    }
    
    void pop() {
        int tmp = s.top();
        s.pop();
        if (tmp >= minValue) {
            int minVal = INT_MAX;
            while (!s.empty()) {
                tmpS.push(s.top());
                minVal 
 
= min(minVal, s.top());
                s.pop();
            }
            minValue = minVal;
        }
        while (!tmpS.empty()) {
            s.push(tmpS.top());
            tmpS.pop();
        }
        
    }
    
    int top() {
        return s.top();
    }
    
    int getMin() {
        return minValue;
    }
    
private:
    stack<int> s;
    stack<int> tmpS;
    int minValue;
};
// 29 ms
/**
 * 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();
 */

[LeetCode] Min Stack