155. 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.

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.

解题思路为two stack。 维护两个stack，一个正常的stack，另一个minStack维护min数据。 每当将x push入stack时，若x <= min，则同时将x push入minStack。 pop时，若minStack和stack值相等，则将minStack中的值也pop出来，需注意的是由于存入的是Integer对象，因此不能用==来判断相等，应使用equal()方法。

public class MinStack {

    /** initialize your data structure here. */
    private Stack<Integer> stack, minStack;
    
    public MinStack() {
        stack = new Stack<>();
        minStack = new Stack<>();
    }
    
    public void push(int x) {
        stack.push(x);
        if(minStack.empty() || x <= minStack.peek()) {
            minStack.push(x);
        }
    }
    
    public void pop() {
        if(stack.pop().equals(minStack.peek())) {
            minStack.pop();
        }
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return minStack.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */