LintCode MinStack

Implement a stack with min() function, which will return the smallest number in the stack.

It should support push, pop and min operation all in O(1) cost.

Example

push(1)
pop()   // return 1
push(2)
push(3)
min()   // return 2
push(1)
min()   // return 1
 

 public class MinStack {
     private Stack<Integer> minStack;
     private Stack<Integer> stack;
     public MinStack() {
         // do initialize if necessary
         minStack = new Stack<Integer>();
         stack = new Stack<Integer>();
     }
 
     public void push(int number) {
         // write your code here
         stack.push(number);
         if (minStack.isEmpty()) {
             minStack.push(number);
         } else {
             if(number <= minStack.peek()) {
                 minStack.push(number);
             }
         }
     }
 
     public int pop() {
         // write your code here
         //here you use equals which stand for two peeked values
         //need to be exactly same.
         //If they are both null, it also works
         if (stack.peek().equals(minStack.peek())) {
             minStack.pop();
         }
         return stack.pop();
     }
 
     public int min() {
         // write your code here
         return minStack.peek();
     }
 }

In this problem, we want to get the current smallest value and implemnt a stack. The main issue is to get current smallest value. So we should use another stack to remember the current smallest values and delete them when we pop that value out.

Trcks: Use stack.peek().equals(minStack.peek()) instead of stack.peek() == minStack.peek() becuase peek() method could return null value and if they are both null it is also okay.