设计一个具有最小值getMin功能的栈设计一个具有最小值getMin功能

设计一个具有getMin功能的栈要求：

1、pop、push、getMin操作时间复杂度都是O(1)
2、设计的栈类型可以使用现有的栈

解法1

import java.util.Stack; public class GetMinStack {private Stack stackData; private Stack stackMin; public GetMinStack() { stackData = https://www.it610.com/article/new Stack<>(); stackMin = new Stack<>(); }public Integer pop() { if (stackData.isEmpty()) { throw new RuntimeException("Stack is Empty!"); } int val = stackData.pop(); if (val == getMin()) { stackMin.pop(); } return val; }public void push(Integer val) { if (stackMin.isEmpty()) { stackMin.push(val); } else if(val <= getMin()){ stackMin.push(val); } stackData.push(val); }public Integer getMin() { if (stackMin.isEmpty()) { throw new RuntimeException("Stack is Empty!"); } return stackMin.peek(); }}

解法2

import java.util.Stack; public class GetMinStack2 {private Stack stackData; private Stack stackMin; public GetMinStack2() { stackData = https://www.it610.com/article/new Stack<>(); stackMin = new Stack<>(); }public Integer pop() { if (stackData.isEmpty()) { throw new RuntimeException("Stack is Empty!"); } stackMin.pop(); return stackData.pop(); }public void push(Integer val) { if (stackMin.isEmpty()) { stackMin.push(val); } else if(val <= getMin()){ stackMin.push(val); } else { int min = stackMin.peek(); stackMin.push(min); } stackData.push(val); }public Integer getMin() { if (stackMin.isEmpty()) { throw new RuntimeException("Stack is Empty!"); } return stackMin.peek(); } }