Day31---學習Java第三彈

阿新 • • 發佈：2021-08-10

2021-08-10

Java經典例題（九）

29、對10個數進行排序

選擇法：

package test2;

public class tset28 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] m = { 1, 5, 4, 6, 7, 4, 5, 8, 56, 45 };
        for (int i = 0; i < m.length; i++) {
            int k = i;
            for 
 (int j = i + 1; j < m.length; j++) {
                if (m[j] < m[k]) {
                    k = j;
                    int temp = m[j];
                    m[j] = m[i];
                    m[i] = temp;
                }
            }
        }
        for (int i = 0; i < m.length; i++) {
            System.out.print(m[i]  
+ " ");
        }
    }

}

冒泡法：

package test2;

public class test28too {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] m = { 1, 5, 4, 6, 7, 4, 5, 8, 56, 45 };
        for (int i = 0; i < m.length; i++) {
            for (int j = 0; j < m.length - 1 - i; j++) {
                 
if (m[j] > m[j + 1]) {
                    int temp = m[j];
                    m[j] = m[j + 1];
                    m[j + 1] = temp;
                }
            }
        }
        for (int i = 0; i < m.length; i++) {
            System.out.print(m[i] + " ");
        }
    }

}

30、求一個3*3矩陣對角線元素之和

package test2;

import java.util.Scanner;

public class test29 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int[][] m = new int[3][3];
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                m[i][j] = sc.nextInt();
            }
        }
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (i == j) {
                    System.out.println(m[i][j]);
                }
            }
        }
    }

}

分析：用雙層for迴圈，如果i==j就輸出

需要注意的點：無

31、有一個已經排好序的陣列。現輸入一個數，要求按原來的規律將它插入陣列中。

package test2;

import java.util.Scanner;

public class test30 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int x = sc.nextInt();
        int[] m = { 1, 2, 6, 9, 12, 34, 56, 78, 90 };
        int[] n = new int[m.length + 1];
        int mark = 0;
        for (int i = 0; i < m.length; i++) {
            if (x > m[i])
                mark = i + 1;
        }
        n[mark] = x;
        for (int i = 0; i < mark; i++) {
            n[i] = m[i];
        }
        for (int i = mark + 1; i < n.length; i++) {
            n[i] = m[i - 1];
        }
        for (int i = 0; i < n.length; i++) {
            System.out.print(n[i] + " ");
        }
    }
}

分析：關鍵是判斷出輸入的數應該在那個位置插入

需要注意的點：注意下標

32、取一個整數a從右端開始的4～7位。

package test2;

import java.util.Scanner;

public class test32 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        char[] ch = s.toCharArray();
        if (ch.length < 7) {
            System.out.println("輸入的數字不符合標準");
            System.exit(-1);
        }

        for (int i = 0; i < ch.length; i++) {
            if (i >= ch.length - 7 && i <= ch.length - 4) {
                System.out.print(ch[i]);
            }
        }
    }

}

分析：用字元陣列解決，簡單直接

需要注意的點：最好判斷一下輸入的數字是否符合要求

33、打印出楊輝三角形（要求打印出10行如下圖）
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

package test2;

public class test33 {
    public static void main(String[] args) {
        int[][] a = new int[10][10];
        for (int i = 0; i < 10; i++) {
            a[i][i] = 1;
            a[i][0] = 1;
        }
        for (int i = 2; i < 10; i++) {
            for (int j = 1; j < i; j++) {
                a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
            }
        }
        for (int i = 0; i < 10; i++) {
            for (int k = 0; k < 2 * (a.length - i) - 1; k++) {
                System.out.print(" ");
            }
            for (int j = 0; j <= i; j++) {
                System.out.print(a[i][j] + "   ");
            }
            System.out.println();
        }
    }
}