leetcode Min Stack题解

题目描述：

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.      

中文理解：实现栈操作，能够在O(1)时间内取到最小的数。

解题思路：采用两个两个栈，同步操作，主栈来正常存储取数据，辅助栈跟着主栈的操作，一样操作，只不过压栈时主栈入数，辅助栈入最小的数。

代码（java）：

class MinStack {

    /** initialize your data structure here. */
    Stack<Integer> mainStack=new Stack<Integer>();
    Stack<Integer> stack=new Stack<Integer>();
    public MinStack() {
        
    }
    
    public void push(int x) {
        mainStack.push(x);
        stack.push(stack.isEmpty() ? x : Math.min(x, stack.peek()));
    }
    
    public void pop() {
        mainStack.pop();
        stack.pop();
    }
    
    public int top() {
        return mainStack.peek();
    }
    
    public int getMin() {
        return stack.peek();
    }
}