Leetcode 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.

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解题思路：

比较直观。用一个min stack专门存放最小值，如果有比它小或是相等的（有多个平行的最小值都要单独存放，否则pop后会出问题），

则存放其到minstack.

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Java code:

class MinStack {
    Stack<Integer> elements = new Stack<Integer>();  
    Stack<Integer> minStack = new Stack<Integer>();  
      
    public void push(int x) {
       elements.push(x);
       if(minStack.isEmpty() || x <= minStack.peek()){
           minStack.push(x);
       }
    }

    public void pop() {
       if(elements.isEmpty()){
           return;
       }
       if((int)elements.peek() == (int)(minStack.peek())){
           minStack.pop();
       }
       elements.pop();
    }

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

    public int getMin() {
        return minStack.peek();
    }
}

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Reference:

1. http://www.cnblogs.com/yuzhangcmu/p/4106783.html