最小栈的实现

public class Code2_Stack_Min { /** * @program: Aglorithm * @Date: today * @Author: Kyrie * @Description: 关于最小栈的实现 * 实现一个栈，该栈带有出栈(pop)、入栈(push)、取最小元素(getMin)3个方法。 * 保证这3个方法的时间复杂度都是O(1)。 * 思路：借助一个辅助栈来实现 */ public Stack<Integer> mainStack = new Stack<>(); //Stack集合 public Stack<Integer> minStack = new Stack<>(); //入栈 public void push(int data){ mainStack.push(data); if(minStack.isEmpty() || data <= minStack.peek()){ //如果辅助栈为空或者其栈顶元素小于要压入栈的元素 minStack.push(data); } } //出栈 public Integer pop() throws Exception{ if (mainStack.isEmpty()){ throw new Exception("stack is empty"); } //若主栈顶元素是最小值，则辅助栈的栈顶先出栈，主栈的栈顶再出栈 if(mainStack.peek().equals(minStack.peek())){ //if(mainStack.peek() == minStack.peek()){ minStack.pop(); } return mainStack.pop(); } //获取最小值 public Integer getMin() throws Exception{ if(minStack.isEmpty()){ throw new Exception("stack is empty"); } return minStack.peek(); } public static void main(String[] args) throws Exception{ Code3_Stack_Min stack_min = new Code3_Stack_Min(); stack_min.push(4); stack_min.push(4); stack_min.push(9); stack_min.push(3); stack_min.push(7); stack_min.push(2); System.out.println(stack_min.minStack.peek()); System.out.println("======================="); System.out.println(stack_min.pop()); System.out.println(stack_min.pop()); System.out.println(stack_min.pop()); System.out.println("======================="); System.out.println(stack_min.minStack.peek()); } } 2 ======================= 2 7 3 ======================= 4