leetcode 155 最小栈

题目：
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：
用一个变量来记录最小值，当新插入的值比之前的最小值要小的时候将旧的值插入栈顶，这样的目的是为了使最小值后面紧跟着次小值，当最小值被弹出时，可以使这个变量马上指向新的最小值

Python
class MinStack(object):

def __init__(self):
    """
    initialize your data structure here.
    """
    self.stack,self.min=[],float('inf')

def push(self, x):
    """
    :type x: int
    :rtype: None
    """
    if(x<=self.min):
        self.stack.append(self.min)
        self.min=x
    self.stack.append(x)
def pop(self):
    """
    :rtype: None
    """
    if(self.stack.pop()==self.min): self.min=self.stack.pop()

def top(self):
    """
    :rtype: int
    """
    return self.stack[-1]
    

def getMin(self):
    """
    :rtype: int
    """
    return self.min

Java
class MinStack {
Stack s;
int min=Integer.MAX_VALUE;
/** initialize your data structure here. */
public MinStack() {
s=new Stack ();
}

public void push(int x) {
    if(x<=min)
    {
        s.push(min);
        min=x;
    }
    s.push(x);
}

public void pop() {
    if(s.pop()==min) min=s.pop();
}

public int top() {
    return s.peek();
}

public int getMin() {
    return min;
}

}