5.Min Stack（能返回最小数的栈）

Level：

  Easy

题目描述：

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.

思路分析：

  使用两个栈，一个栈压入元素，另一个栈栈顶始终保存已经入栈元素中的最小值。

代码：

class MinStack {

    /** initialize your data structure here. */
    Stack<Integer>pushStack=new Stack<>();
    Stack<Integer>minStack=new Stack<>();
    public MinStack() {
        
    }
    
    public void push(int x) {
        pushStack.push(x);
        if(minStack.isEmpty()||x<=minStack.peek()){
            minStack.push(x);
        }
    }
    
    public void pop() {
        if(!pushStack.isEmpty()){
        if((int)pushStack.peek()==(int)minStack.peek()){
            minStack.pop();
        }
        pushStack.pop();
        }
    }
    
    public int top() {
        if(!pushStack.isEmpty())
        return pushStack.peek();
        else
            return -1;
    }
    
    public int getMin() {
        return minStack.peek();
    }
}