编写一个程序以按升序对堆栈排序[英] Write a program to sort a stack in ascending order

问题描述

有人可以帮忙看看我的代码吗?非常感谢你的帮助.输入堆栈是 [5, 2, 1, 9, 0, 10],我的代码给出了输出堆栈 [0, 9, 1, 2, 5, 10],9 不在正确的位置.

import java.util.*;

public class CC3_6 {
public static void main(String[] args) {
    int[] data = {5, 2, 1, 9, 0, 10};
    Stack<Integer> myStack = new Stack<Integer>();
    for (int i = 0; i < data.length; i++){
        myStack.push(data[i]);
    }
    System.out.println(sortStack(myStack));
}

public static Stack<Integer> sortStack(Stack<Integer> origin) {
    if (origin == null)
        return null;
    if (origin.size() < 2)
        return origin;

    Stack<Integer> result =  new Stack<Integer>();
    while (!origin.isEmpty()) {
        int smallest = origin.pop();
        int remainder = origin.size();
        for (int i = 0; i < remainder; i++) {
            int element = origin.pop();
            if (element < smallest) {
                origin.push(smallest);
                smallest = element;                    
            } else {
                origin.push(element);
            }
        }
        result.push(smallest);
    }
    return result;

}

}

推荐答案

package TwoStackSort;
import java.util.Random;
import java.util.Stack;

public class TwoStackSort {
    /**
     *
     * @param stack1 The stack in which the maximum number is to be found.
     * @param stack2 An auxiliary stack to help.
     * @return The maximum integer in that stack.
     */
    private static Integer MaxInStack(Stack<Integer> stack1, Stack<Integer> stack2){
        if(!stack1.empty()) {
            int n = stack1.size();
            int a = stack1.pop();
            for (int i = 0; i < n-1; i++) {
                if(a <= stack1.peek()){
                    stack2.push(a);
                    a = stack1.pop();
                }
                else {
                    stack2.push(stack1.pop());
                }
            }
            return a;
        }
        return -1;
    }

    /**
     *
     * @param stack1 The original stack.
     * @param stack2 The auxiliary stack.
     * @param n An auxiliary parameter to keep a record of the levels of recursion.
     */
    private static void StackSort(Stack<Integer> stack1, Stack<Integer> stack2, int n){
        if(n==0){
            return;
        }
        else{
            int maxinS1 = MaxInStack(stack1, stack2);
            StackSort(stack2, stack1, n-1);
            if(n%2==0){
                stack2.push(maxinS1);
            }
            else{stack1.push(maxinS1);}
        }
    }
    /**
     *
      * @param stack1 The original stack that needs to be sorted.
     * @param stack2 The auxiliary stack.
     * @return The descendingly sorted stack.
     */
    public static Stack<Integer> TwoStackSorter(Stack<Integer> stack1, Stack<Integer> stack2){
        StackSort(stack1, stack2, stack1.size()+stack2.size());
        return (stack1.empty())? stack2:stack1;
    }

    public static void main(String[] args) {
        Stack<Integer> stack = new Stack<>();
        Random random = new Random();
        for (int i = 0; i < 50; i++) {
            stack.push(random.nextInt(51));
        }
        System.out.println("The original stack is: ");
        System.out.print(stack);
        System.out.println("\n" + "\n");
        Stack<Integer> emptyStack = new Stack<>();
        Stack<Integer> res =  TwoStackSorter(stack, emptyStack);
        System.out.println("The sorted stack is: ");
        System.out.print(res);
    }
}

这是我昨天晚上经过一个小时的头脑风暴后想出的代码.当我解决这个问题的一个版本时,我有一个限制,最多只能使用一个额外的堆栈.这是这个问题的一个强烈的递归解决方案.我使用了 2 种私有方法从堆栈中获取我需要的东西.我真的很喜欢递归在这里的工作方式.基本上,我正在解决的版本需要通过最多使用一个额外的堆栈以升序/降序对堆栈进行排序.请注意,不应使用其他数据结构.

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