Question

我目前正在学习Java。下面是我编写的一个简单Java程序的方法列表。这些方法中有什么突出的东西会导致程序执行速度变慢吗？使用仅包含6个整数的数组执行时间为4秒：

EDITED：按照要求提供整个程序。我在Textpad中写了它。我意识到它不是最有效的算法。它完成它应该做的事情，但需要很长时间才能完成。

import java.util.*;
public class Supermarket
{
    public static void main(String [] args)
    {
        int[] custTimes =
        {
            1, 6, 7, 4, 4, 3, 5, 1, 2, 1, 3, 6, 4
        };

        int checkOuts = 6;
        int answer;
        answer = Solution.solveSuperMarketQueue(custTimes, checkOuts);
        System.out.println("Answer is " + answer);
    }
}//~public class Supermarket...

class Solution
{
    static int myTotal;
    static int solveSuperMarketQueue(int[] customers, int n)
    {
        // ******************* INITIALIATION ***********************
        myTotal = 0;
        int len = customers.length;  // length of customer queue
        if (len < 1)
        {
            return 0;
        }
        int[] till = new int[n]; // array to store all tills and till queues
        int tillMin; // Minimum time
        int tillMax; // Maximum time

        // Put the customers into an arraylist:
        ArrayList<Integer> times = new ArrayList<Integer>();
        for (int i = 0; i < len; i = i + 1)
        {
            times.add(i, customers[i]);
        }

        // create the array of tills and set all queue intial values to 0
        for (int i = 0; i < n; n = n + 1)
        {
            till[i] = 0;
        }
        // Move the queue to tills to start off
        ReturnPair result = copyQueue(till, times);
        till = result.getArr();
        times = result.getList();
        int s = times.size();

        tillMax = getMaxTime(till);
        tillMin = getMinTime(till);

        // ***************** END OF INITIALIATION ******************

        // *****************MAIN LOOP ******************************


        while (tillMax > 0)
        {
            // Find customer(s) with least time use  that time to move all queues
            // and update myTotal time.
            // STEP 1: get minimum time in tills array (ignore zero)
            tillMin = getMinTime(till);
            // STEP 2: subtract minimum value from all the tills, but not if till has a zero
            if (tillMin > 0)
            {
                till = subtractTime(till, tillMin);
            }
            // Move the queue to tills
            if (s > 0)
            {
                result = copyQueue(till, times);
                till = result.getArr();
                times = result.getList();
            }

            tillMax = getMaxTime(till);
            tillMin = getMinTime(till);
        }
        return myTotal;

        // **************** END OF LOOP *****************************

    }//~public static int solveS...


    // ****************** METHODS **********************************

    // Method to move queue foward
    // For each till, a time is copied from the customer array.
    // The values are copied in order.
    // The value is coped only if array value is zero.
    private static ReturnPair copyQueue(int[] arr, ArrayList<Integer> arrList)
    {
        int n = arr.length; // for each till...
        for (int i = 0; i < n; i = i + 1)
        {
            if (arr[i] == 0 && arrList.size() > 0) // only copy if it current till value is 0 AND arrayList value exists
            {
                arr[i] = arrList.get(0);
                arrList.remove(0);
            }
        }

        // returns an instance of the object myResult which is a container for an array and an arraylist
        return new ReturnPair(arr, arrList);
    }

    // Method to get minimum time from array (but not zero).
    private static int getMinTime(int[] arr)
    {
        int minValue = 0;

        // make sure arr[i] isn't zero.
        for (int i = 0; i < arr.length; i = i + 1)
        {
            if (arr[i] != 0)
            {
                minValue = arr[i];
                break;
            }
        }

        // Find minimum value that isn't zero.
        for (int i = 1; i < arr.length; i = i + 1)
        {
            if (arr[i] != 0 && arr[i] < minValue)
            {
                minValue = arr[i];
            }
        }
        return minValue;
    }//~static int getMinTime(in...

    // Method to subtract minimum time from tills
    private static int[] subtractTime(int[] arr, int min)
    {
        int n = arr.length;
        for (int i = 0; i < n; i = i + 1)
        {
            if (arr[i] != 0)
            {
                arr[i] = arr[i] - min;
            }
        }
        // update myTotal
        myTotal = myTotal + min;
        return arr;
    }//~static void subtractTime...

    private static int getMaxTime(int[] arr)
    {
        int maxValue = arr[0];
        for (int i = 1; i < arr.length; i = i + 1)
        {
            if (arr[i] > maxValue)
            {
                maxValue = arr[i];
            }
        }
        return maxValue;
    }
}//~class Solution...

// Special class designed to return an array and an array list as an object
class ReturnPair
{
    // set up fields
    int[] newArr;
    ArrayList<Integer> newArrList;

    // define method
    public ReturnPair(int[] first, ArrayList<Integer> second)
    {
        this.newArr = first;
        this.newArrList = second;
    }

    public int[] getArr()
    {
        return newArr;
    }

    public ArrayList<Integer> getList()
    {
        return newArrList;
    }
}

Answer 1

for (int i = 0; i < n; n = n + 1)

此行正在递增n而不是i。它将循环直到n溢出。它应该是：

for (int i = 0; i < n; i++)

因为int数组无论如何都被初始化为0，所以你可以完全删除这个循环。

Java方法定义相当缓慢执行

1 个答案: