转载 关于算法的五个有趣的问题

问题1

使用for循环、while循环和递归写出3个函数来计算给定数列的总和。

package com.luka;

public class Javat {

    private static int[] arr_Ints = { 2, 1, 4, 3, 6, 5, 8, 7, 10, 9 };

    public static void main(String[] args) {
        System.out.println("The Count is " + getNumByFor() + " .");
        System.out.println("The Count is " + getNumByWhile() + " .");
        System.out.println("The Count is " + getNumByEcursion(0) + " .");
    }

    /**
     * for 循环
     */
    private static int getNumByFor() {
        int count = 0;
        for (int i = 0; i < arr_Ints.length; i++) {
            count += arr_Ints[i];
        }
        return count;
    }

    /**
     * while 循环
     */
    private static int getNumByWhile() {
        int count = 0;
        int i = 0;
        while (i < arr_Ints.length) {
            count += arr_Ints[i];
            i++;
        }
        return count;
    }

    /**
     * 递归
     */
    private static int getNumByEcursion(int i) {
        if (arr_Ints.length == 0)
            return 0;
        else if (i < arr_Ints.length - 1)
            return arr_Ints[i] + getNumByEcursion(i + 1);
        else
            return arr_Ints[i];
    }
}

问题2

编写一个交错合并列表元素的函数。例如：给定的两个列表为[a，B，C]和[1，2，3]，函数返回[a，1，B，2，C，3]。

package com.luka;

public class Javat {

    private static String[] arr1 = { "a", "B", "c", "D", "e" };
    private static String[] arr2 = { "1", "2", "3" };

    public static void main(String[] args) {
        String[] arr = getNum(arr1, arr2);
        for (int i = 0; i < arr.length; i++)
            System.out.println("The Num is " + arr[i] + " .");
    }

    private static String[] getNum(String[] arr12, String[] arr22) {
        String[] arr = new String[arr1.length + arr2.length];
        int i, j;
        for (i = 0, j = 0; i < arr1.length; i++) {
            j = 2 * i;
            if (j > 2 * arr2.length)
                j = arr2.length + i;
            arr[j] = arr1[i];
        }
        for (i = 0, j = 0; i < arr2.length; i++) {
            j = 2 * i + 1;
            if (j > 2 * arr1.length)
                j = arr1.length + i;
            arr[j] = arr2[i];
        }
        return arr;
    }

}

问题3

编写一个计算前100位斐波那契数的函数。根据定义，斐波那契序列的前两位数字是0和1，随后的每个数字是前两个数字的和。例如，前10位斐波那契数为：0，1，1，2，3，5，8，13，21，34。

package com.luka;

public class Javat {
    public static void main(String[] args) {
        try {
            System.out.println("The Nums is " + getCount(100) + " .");
        } catch (Exception e) {
        }
    }

    // 获取值
    private static int getNum(int num) {
        int count = 0;
        if (num <= 1)
            count = 0;
        else if (num == 2)
            count = 1;
        else
            count = getNum(num - 1) + getNum(num - 2);
        return count;
    }

    // 获取和
    private static String getCount(int num) {
        String strNums = "";
        for (int i = 0; i <= num; i++) {
            strNums += getNum(i) + ",";
        }
        strNums = strNums.substring(0, strNums.length()-1);
        return strNums;
    }

}

问题4

编写一个能将给定非负整数列表中的数字排列成最大数字的函数。例如，给定[50，2，1,9]，最大数字为95021。

package com.luka;

import java.util.Arrays;
import java.util.Comparator;

public class Javat {

    private static Integer[] VALUES = { 50, 2, 100, 99, 5, 7, 51,50 ,11};

    public static void main(String[] args) {
        Arrays.sort(VALUES, new Comparator<Object>() {
            @Override
            public int compare(Object lhs, Object rhs) {
                String v1 = lhs.toString();
                String v2 = rhs.toString();

                return (v1 + v2).compareTo(v2 + v1) * -1;
            }
        });

        String result = "";
        for (Integer integer : VALUES) {
            result += integer.toString();
        }

        System.out.println(result);
    }

}

问题5

编写一个在1，2，…，9（顺序不能变）数字之间插入+或-或什么都不插入，使得计算结果总是100的程序，并输出所有的可能性。例如：1 + 2 + 34 – 5 + 67 – 8 + 9 = 100。

package com.luka;

import java.util.ArrayList;

public class Javat {
    private static int TARGET_SUM = 100;
    private static int[] VALUES = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    private static ArrayList add(int digit, String sign, ArrayList branches) {
        for (int i = 0; i < branches.size(); i++) {
            branches.set(i, digit + sign + branches.get(i));
        }

        return branches;
    }

    private static ArrayList f(int sum, int number, int index) {
        int digit = Math.abs(number % 10);
        if (index >= VALUES.length) {
            if (sum == number) {
                ArrayList result = new ArrayList();
                result.add(Integer.toString(digit));
                return result;
            } else {
                return new ArrayList();
            }
        }

        ArrayList branch1 = f(sum - number, VALUES[index], index + 1);
        ArrayList branch2 = f(sum - number, -VALUES[index], index + 1);

        int concatenatedNumber = number >= 0 ? 10 * number + VALUES[index] : 10
                * number - VALUES[index];
        ArrayList branch3 = f(sum, concatenatedNumber, index + 1);

        ArrayList results = new ArrayList();

        results.addAll(add(digit, "+", branch1));
        results.addAll(add(digit, "-", branch2));
        results.addAll(add(digit, "", branch3));

        return results;
    }

    public static void main(String[] args) {
        ArrayList list = f(TARGET_SUM, VALUES[0], 1);
        for(int i=0;i<list.size();i++)
        {            
            System.out.println(list.get(i));            
        }
    }

}

输出结果：

1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89