封装一个方法sum,输入一个非负整数,求1 ~ 一个数之间的和,并返回
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int a = scan.nextInt(); int sum = sum(a); System.out.println(sum); } public static int sum(int num) { int sum = 0; for (int i = 1; i <= num; i++) { sum+=i; } return sum; } }
封装一个方法sum,输入两个非负整数,求两个数之间的和,并返回 (提升)
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int a = scan.nextInt(); int b = scan.nextInt(); int sum = sum(a,b); System.out.println(sum); } public static int sum(int num1,int num2) { int sum = 0; int max = num1>num2?num1:num2; int min = num1最大值sum+= i; } return sum; } }
封装一个方法max2,求两个数的最大值,并返回
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int a = scan.nextInt(); int b = scan.nextInt(); int max = max2(a, b); System.out.println(max); } public static int max2(int num1,int num2) { return num1>num2?num1:num2; } }
封装一个方法max3,求三个数的最大值,并返回(提升)
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方法一:
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int a = scan.nextInt(); int b = scan.nextInt(); int c = scan.nextInt(); int max = max3(a, b, c); System.out.println(max); } public static int max3(int num1,int num2,int num3) { return num1>num2?(num1>num3?num1:num3):(num2>num3?num2:num3); } }
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方法二:
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int a = scan.nextInt(); int b = scan.nextInt(); int c = scan.nextInt(); int max = max2(max2(a, b),c); System.out.println(max); } public static int max2(int num1,int num2) { return num1>num2?num1:num2; } }
封装一个sum方法,给定一个四位的数字,求各个位数之和
public class Demo { public static void main(String[] args) { sum1(1234); } public static void sum1(int num){ int sum = 0;//和 while (num > 0 ){ sum+=num%10; num=num/10; } System.out.println("各位数字之和: " + sum); } }
封装一个sum方法,给定一个四位的数字,求各个位数之和,并返回
public class Demo { public static void main(String[] args) { //调用方法 sum2 int result = sum2(1234); System.out.println(result); } public static int sum2(int num){ int sum = 0;//和 while (num > 0 ){ sum+=num%10; num=num/10; } return sum; } }敲7
封装一个方法,打印出1 ~ 100 之间所有不是7的倍数和不包含7的数字,并求和
public class Demo { public static void main(String[] args) { //调用方法 sum1 sum1(); } public static void sum1(){ int sum = 0;//和 for(int i = 1; i <= 100; i++) { // i%7 == 0 表示 7的倍数 // i/10 == 7 表示十位数上的数字为7 // i%10 == 7 表示个位数上的数字为7 if (i % 7 == 0 || i / 10 == 7 || i % 10 == 7){ continue; } System.out.print(i + " "); sum += i; } System.out.println("和为: " + sum); } }
封装一个方法,求1 ~ 100 之间所有不是7的倍数和不包含7的数字之和 ,将该值返回
public class Demo { public static void main(String[] args) { //调用方法 sum2 int result = sum2(1, 100); System.out.println("和为: " + result); } public static int sum2(){ int sum = 0;//和 for(int i = 1; i <= 100; i++) { // i%7 == 0 表示 7的倍数 // i/10 == 7 表示十位数上的数字为7 // i%10 == 7 表示个位数上的数字为7 if (i % 7 == 0 || i / 10 == 7 || i % 10 == 7){ continue; } sum += i; } return sum; } }数组的操作
封装一个generate方法,用于生成指定长度的int数组,数组的元素为0到指定范围的随机数,并将该数组返回
public class Demo { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int len = scan.nextInt(); int max = scan.nextInt(); int[] arr = generate(len,max); System.out.println(Arrays.toString(arr)); } public static int[] generate(int len,int max) { int[] arr = new int[len]; for (int i = 0; i < arr.length; i++) { arr[i] = (int)(Math.random()*(max+1)); } return arr; } }
给出两个数组:
- 数组a:{1,7,9,11,13,15,17,19};
- 数组b:{2,4,6,8,10}
封装一个方法mergeArr将两个数组合并为数组c,按升序排列,并将数组返回。
public class Demo { public static void main(String[] args) { int[] arr1 = {1,7,9,11,13,15,17,19}; int[] arr2 = {2,4,6,8,10}; int[] arr = mergeArr(arr1,arr2); System.out.println(Arrays.toString(arr)); } public static int[] mergeArr(int[] arr1,int[] arr2) { int[] arr = new int[arr1.length+arr2.length]; System.arraycopy(arr1,0,arr,0,arr1.length); System.arraycopy(arr2,0,arr,arr1.length,arr2.length); Arrays.sort(arr); return arr; } }