min-stack

155. Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

MinStack() initializes the stack object.
void push(int val) pushes the element val onto the stack.
void pop() removes the element on the top of the stack.
int top() gets the top element of the stack.
int getMin() retrieves the minimum element in the stack.

You must implement a solution with O(1) time complexity for each function.

Example 1:

Input ["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]]

Output [null,null,null,null,-3,null,0,-2]

Explanation 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

Constraints:

-231 <= val <= 231 - 1
Methods pop, top and getMin operations will always be called on non-empty stacks.
At most 3 * 104 calls will be made to push, pop, top, and getMin.

class MinStack {
public:
    MinStack() {}

    void push(int val) {
        s1.push(val);
        if(s2.empty() || val <= s2.top()) {
            s2.push(val);
        }
    }

    void pop() {
        // 如果 s1.top() == 最小值的話
        if(s1.top() == s2.top()) {
            s2.pop();
        }
        s1.pop();
    }

    int top() {
        return s1.top();
    }

    int getMin() {
        return s2.top();
    }
private:
    // 用兩個 stack 存
    // s1 -> 一般的 stack
    // s2 -> 用來存最小值的 stack
    stack<int> s1;
    stack<int> s2;
};

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

T: $O(1)$
S: $O(N)$

155. Min Stack​

155. Min Stack