java 要求用时空复杂度O(1)求出栈（结构自定）中的最大最小值

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package com.zhangheng;

import java.util.Stack;

/**
 * 栈的使用  高效求栈中的最大最小值 时空复杂度均是O(1)
 * @date 2018-10-26 
 * @author [email protected]
 *
 * @param <E>
 */
public class SpecialStack<E extends Number> {
	Stack<Long> stack = new Stack<Long>();
	Stack<Long> stackMin = new Stack<Long>();// 存放求最小值的栈
	Stack<Long> stackMax = new Stack<Long>();// 存放求最大值的栈

	public void push(Long l) {
		stack.push(l);

		if(stackMin.isEmpty()){
			stackMin.push(l);
		}else{
			if(l.longValue() <= stackMin.peek()){
				stackMin.push(l);
			}
		}
		
		
		if(stackMax.isEmpty()){
			stackMax.push(l);
		}else{
			if(l.longValue() > stackMax.peek()){
				stackMax.push(l);
			}
		}
		
	}

	public Long pop()// 一定要记着，非空才能pop()
	{
		if (!stack.isEmpty() && !stackMin.isEmpty() && !stackMax.isEmpty()) {
			Long e = stack.pop();
			stackMin.pop();
			stackMax.pop();
			return e;
		} else {
			System.out.println("栈为空了");
			return null;
		}

	}

	public Long getMin() {
		return stackMin.peek();
	}

	public Long getMax() {
		return stackMax.peek();
	}

	
	public static void main(String[] args) {
		SpecialStack<Long> spect = new SpecialStack<Long>();
		spect.push(2l);
		spect.push(1l);
		spect.push(4l);
		spect.push(7l);
		spect.push(5l);
		
		
		System.out.println("spect.stack1="+spect.stack);
		System.out.println("spect.stackMin="+spect.stackMin);
		System.out.println("spect.getMin()="+spect.getMin());
		System.out.println("spect.stackMax="+spect.stackMax);
		System.out.println("spect.getMax()="+spect.getMax());
	}
}

输出结果：
spect.stack1=[2, 1, 4, 7, 5]
spect.stackMin=[2, 1]
spect.getMin()=1
spect.stackMax=[2, 4, 7]
spect.getMax()=7