20162330 第三周 蓝墨云班课 算法复杂度练习

题目:用 Java语言实现下列算法并进行单元测试, 请给出算法的时间复杂度。

  • (1)求一个整数二维数组Arr[N][N]的所有元素之和。
    (2)对于输入的任意 3 个整数, 将它们按从小到大的顺序输出。
    (3)对于输入的任意 n 个整数, 输出其中的最大和最小元素。

(1)求一个整数二维数组Arr[N][N]的所有元素之和。

  • 代码如下:
public class Test1 {
        public static void main(String[] args) {
            int[][] nums = { { 10, 20 }, { 5, 25 }, { 1, 29 } };
            System.out.println(getSum(nums));
        }

        public static int getSum(int[][] arrs) {
            int sum = 0;
            for (int i = 0; i < arrs.length; i++) {
                for (int j = 0; j < arrs[i].length; j++) {
                    sum += arrs[i][j];
                }
            }
            return sum;
        }
}

单元测试:

public class Test1Test {
    /**
     * Method: getSum(int[][] arrs)
     */
    @Test
    public void testGetSum() throws Exception {
        int[][] list1 = new int[][]{{10, 5}, {25, 25}, {6, 29}};
        Assert.assertEquals(100, Test1.getSum(list1));
        int[][] list2 = new int[][]{{10, 5}, {25, 25}};
        Assert.assertEquals(65, Test1.getSum(list2));
        int[][] list3 = new int[][]{{25, 25}};
        Assert.assertEquals(50, Test1.getSum(list3));
    }
} 

复杂度分析】这个例子我给出了二维数组的具体元素,如果没有给出具体元素,即有n个含有n个元素的一维数组,遍历求和要用for的嵌套循环,所以时间复杂度为 O(\(n^2\)) .


(2)对于输入的任意 3 个整数, 将它们按从小到大的顺序输出。

  • 代码如下:
public class Test2 {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        System.out.println("请输入第一个整数:");
        int a = scan.nextInt();

        System.out.println("请输入第二个整数:");
        int b = scan.nextInt();

        System.out.println("请输入第三个整数:");
        int c = scan.nextInt();

        Compare(a, b, c);
    }

    public static String Compare(int a, int b, int c) {
        int t;
        if (a == b || a == c || b == c)
            return "Equality exist.";
        else {
            if (a < b) {
                t = b;
                b = a;
                a = t;
            }
            if (a < c) {
                t = c;
                c = a;
                a = t;
            }
            if (b < c) {
                t = c;
                c = b;
                b = t;
            }
        }
        return c + " < " + b + " < " + a;
    }
}

单元测试:

public class Test2Test {
    /**
     * Method: Compare(int a, int b, int c)
     */
    @Test
    public void testCompare() throws Exception {
        int a1 = 1, b1 = 2, c1 = 3;
        Assert.assertEquals("1 < 2 < 3", Test2.Compare(a1, b1, c1));
        int a2 = 20, b2 = 30, c2 = 10;
        Assert.assertEquals("10 < 20 < 30", Test2.Compare(a2, b2, c2));
        int a3 = 22, b3 = 22, c3 = 11;  // bound test
        Assert.assertEquals("Equality exist.", Test2.Compare(a3, b3, c3));
    }
} 

复杂度分析】这个例子通过交换赋值法进行排序,规定输入3个整数,算法执行次数确定,所以时间复杂度为 O(1) .


(3)对于输入的任意 n 个整数, 输出其中的最大和最小元素。

  • 代码如下:
public class Test3 {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        System.out.println("请输入四个整数:");
        int a = scan.nextInt();
        int b = scan.nextInt();
        int c = scan.nextInt();
        int d = scan.nextInt();
        Comparable[] arr = {a, b, c, d};
        selectionSort(arr);
    }

    public static String selectionSort(Comparable[] data) {
        int min;

        for (int index = 0; index < data.length - 1; index++) {
            min = index;
            for (int scan = index + 1; scan < data.length; scan++)
                if (data[scan].compareTo(data[min]) < 0)
                    min = scan;
            swap(data, min, index);
        }
        return "Max:" + data[3] + ",Min:" + data[0];
    }

    private static void swap(Comparable[] data, int index1, int index2) {
        Comparable temp = data[index1];
        data[index1] = data[index2];
        data[index2] = temp;
    }
}

单元测试:

public class Test3Test {
    /**
     * Method: selectionSort(Comparable[] data)
     */
    @Test
    public void testSelectionSort() throws Exception {
        Comparable[] list1 = {11, 22, 33, 44};
        Assert.assertEquals("Max:44,Min:11", Test3.selectionSort(list1));
        Comparable[] list2 = {55, 66, 33, 44};
        Assert.assertEquals("Max:66,Min:33", Test3.selectionSort(list2));
        Comparable[] list3 = {44, 44, 44, 44};  // bound test
        Assert.assertEquals("Max:44,Min:44", Test3.selectionSort(list3));
    }
} 

复杂度分析】这个例子我只添加了4个整数,如果输入n个整数,使用选择排序(for嵌套循环)再返回,所以时间复杂度为 O(\(n^2\)) .

posted @ 2017-09-24 15:53  N-Liu  阅读(743)  评论(0编辑  收藏  举报