Daily practice------There are n integers, so that the previous numbers are moved back m positions in order, and the last m numbers become the first m numbers

题目: 有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数

解题关键:It is necessary to create a new array so that the original array can be moved backward as a wholeM位

思路:1.创建个有n个整数的数组

        2.输出n个数字,存储到数组中

        3.遍历原数组

        4.The sequence of the previous numbers in the output is shifted backwardm个位置

        5.Create a new array of length (n+m)

        6.Store the values ​​of the old array into the new array

        7.move the whole array backwardsm位

        8.m个数变成最前面的m个数

        9.遍历新数组

过程: 接下来我们根据我们的解题思路来一步步写代码

        1.创建个有n个整数的数组
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter how many integers you need to enter:");
        int n = sc.nextInt();
        int[] nums = new int[n];

        2.输出n个数字,存储到数组中
        System.out.println("请依次输入" + n + "个整数");
        for (int i = 0; i < nums.length; i++) {
            System.out.println("第" + (i + 1) + "个数字为:");
            int num = sc.nextInt();
            nums[i] = num;
        }

         3.遍历原数组
        System.out.print("原数组为:");
        for (int i = 0; i < nums.length; i++) {
            System.out.print(nums[i] + " ");
        }

        System.out.println();// 换行

         4.The sequence of the previous numbers in the output is shifted backwardm个位置
        System.out.print("Output the position where the preceding numbers are sequentially moved backwards:");
        int m = sc.nextInt();

        5.Create a new array of length (n+m)
        int[] newNums = new int[n + m];
        
        6.Store the values ​​of the old array into the new array
        for (int i = 0; i < nums.length; i++) {
            newNums[i] = nums[i];
        }
        System.out.println();
        7.move the whole array backwardsm位
        for (int i = newNums.length - 1; i-m >= 0 ; i--) {
            newNums[i] = newNums[i-m];
        }
        
        8.m个数变成最前面的m个数
        for (int i = 0; i < m; i++) {
            newNums[i] = newNums[newNums.length -(m-i)];
        }
        
        9.遍历新数组
        System.out.print("新数组为:");
        for (int i = 0; i < nums.length; i++) {
            nums[i] = newNums[i];
            System.out.print(nums[i] + " ");
        }

完整结果如下:

   为了方便大家使用,下面附上源码:

		// 1.创建个有n个整数的数组
		Scanner sc = new Scanner(System.in);
		System.out.println("Please enter how many integers you need to enter:");
		int n = sc.nextInt();
		int[] nums = new int[n];

		// 2.输出n个数字,存储到数组中
		System.out.println("请依次输入" + n + "个整数");
		for (int i = 0; i < nums.length; i++) {
			System.out.println("第" + (i + 1) + "个数字为:");
			int num = sc.nextInt();
			nums[i] = num;
		}

		// 3.遍历原数组
		System.out.print("原数组为:");
		for (int i = 0; i < nums.length; i++) {
			System.out.print(nums[i] + " ");
		}

		System.out.println();// 换行

		// 4.The sequence of the previous numbers in the output is shifted backwardm个位置
		System.out.print("Output the position where the preceding numbers are sequentially moved backwards:");
		int m = sc.nextInt();

		// 5.Create a new array of length (n+m)
		int[] newNums = new int[n + m];
		
		//6.Store the values ​​of the old array into the new array
		for (int i = 0; i < nums.length; i++) {
			newNums[i] = nums[i];
		}
		System.out.println();
		//7.move the whole array backwardsm位
		for (int i = newNums.length - 1; i-m >= 0 ; i--) {
			newNums[i] = newNums[i-m];
		}
		
		//8.m个数变成最前面的m个数
		for (int i = 0; i < m; i++) {
			newNums[i] = newNums[newNums.length -(m-i)];
		}
		
		//9.遍历新数组
		System.out.print("新数组为:");
		for (int i = 0; i < nums.length; i++) {
			nums[i] = newNums[i];
			System.out.print(nums[i] + " ");
		}

明日练习:定义一个N*N二维数组,从键盘上输入值,找出每行中最大值组成一个一维数组并输出;

大家可以自己写写,明天中午12点我准时发出我的写法哦,明天12点不见不散

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