155. Min Stack

Description:

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.

Solutions:

Approach: Use two stacks

用两个栈，一个栈stackData用作普通的栈存放数据，一个栈stackMin只存放当前最小的值。
注意两点：

1. 如果当前入栈的数等于stackMin栈顶的数，则该数两个栈都要压入。弹栈时，若两个栈顶的数大小相同，则都弹出。
2. 比较栈顶的数时不能用等于号（==），因为栈内元素为包装过的Integer对象，若要比较要通过intValue()方法获取int值后再比较。
   代码如下：

import java.util.Stack;
class MinStack {

    /** initialize your data structure here. */
    private Stack stackData;
    private Stack stackMin;
    public MinStack() {
        stackData = new Stack();
        stackMin = new Stack();
    }
    
    public void push(int x) {
        if (stackData.isEmpty() || x <= stackMin.peek()) {
            stackMin.push(x);
        }
            stackData.push(x);
    }
    
    public void pop() {
        if (stackData.isEmpty()) {
            return;
        }
        int val = stackData.pop();
        if (stackMin.peek().intValue() == val) {
            stackMin.pop();
        } 
    }
    
    public int top() {
        return stackData.peek();
    }
    
    public int getMin() {
        return stackMin.peek();
    }
}