通过接受字符串使用for循环打印模式(printing a pattern using for loop by accepting a string)

通过接受字符串使用for循环打印模式(printing a pattern using for loop by accepting a string)

示例输入字符串：

BlueJ的

输出：

b***b *l*l* **u** *e*e* j***j

（其中'*'表示空格）

到目前为止我做了这么多：

int n = s.length() - 1; int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= (n + 1); j++) { if (i == j || i == n - (j - 1)) System.out.print(s.charAt(i)); else { System.out.print("*"); } } System.out.println(); }

但输出是：

b****b *l**l* **uu** **ee** *j**j*

Sample input string:

bluej

Output:

b***b *l*l* **u** *e*e* j***j

(where '*' indicates a blank space)

I have done this much so far:

int n = s.length() - 1; int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= (n + 1); j++) { if (i == j || i == n - (j - 1)) System.out.print(s.charAt(i)); else { System.out.print("*"); } } System.out.println(); }

But output is:

b****b *l**l* **uu** **ee** *j**j*

最满意答案

只需将i == n-(j-1)替换为i == n - j ，将第二个替换for (j = 0; j <= n+1; j++)循环for (j = 0; j <= n; j++) ：

int n = s.length() - 1; int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= n; j++) { if (i == j || i == n - j) System.out.print(s.charAt(i)); else { System.out.print("*"); } } System.out.println(); }

Just replace i == n-(j-1) with i == n - j and second for loops for (j = 0; j <= n+1; j++) to for (j = 0; j <= n; j++) :

int n = s.length() - 1; int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= n; j++) { if (i == j || i == n - j) System.out.print(s.charAt(i)); else { System.out.print("*"); } } System.out.println(); }