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.

这题很简单，主要是有有一个需要注意的地方，就是对于对象内容的比较，不能用基本类型，需要用基本类型的包装类，不能用==，要使用equals方法。在这个问题上纠结了好久，在网上找到了答案 参考 点击打开链接

public class MinStack {

    /** initialize your data structure here. */
    
    private Stack<Integer> stack=new Stack();
    private Stack<Integer> min_stack=new Stack();
 
    
    public void push(int x) {
        stack.push(x);
        if(min_stack.empty() || x<=min_stack.peek())  //把当前最小的值进栈
        {
            min_stack.push(x);
        }
    }
    
    public void pop() {
       
        if(stack.peek().equals(min_stack.peek())){
            min_stack.pop();
        }
        stack.pop();
        
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return min_stack.peek();
    }
}

/**
 * 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();
 */