[LeetCode] 155. Min Stack

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.

class MinStack {
public:
    /** initialize your data structure here. */
    MinStack() {
        
    }
    
    void push(int x) {
        m.push(x);
        if (h.size() == 0 || h.top() >= x) {
            h.push(x);
        } else {
            h.push(h.top());
        }
    }
    
    void pop() {
        m.pop();
        h.pop();
    }
    
    int top() {
        return m.top();
    }
    
    int getMin() {
        return h.top();
    }
private:
    stack<int> m;
    stack<int> h;
};

/**
 * 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();
 */

 解析：以上解法用了一个辅助栈，还可以不使用辅助栈，用一个变量来存储最小值。min_val来记录当前最小值，初始化为整型最大值，然后如果需要进栈的数字小于等于当前最小值min_val，那么将min_val压入栈，

并且将min_val更新为当前数字。在出栈操作时，先将栈顶元素移出栈，再判断该元素是否和min_val相等，相等的话我们将min_val更新为新栈顶元素，再将新栈顶元素移出栈即可。