原文:
给定奇数长度的字符串，以 x 格式打印该字符串。 例:
input: 12345
output:
1 5
2 4
3
2 4
1 5
input: geeksforgeeks
output:
g s
e k
e e
k e
s g
f r
o
f r
s g
k e
e e
e k
g s
我们强烈建议你尽量减少浏览器，先自己试试这个。 想法是在单个循环中使用两个变量，第一个变量‘i’从左到右，第二个变量‘j’从右到左。十字(或 x)的上半部分在它们相遇之前被打印出来。中心字符在它们相遇时被打印，下分字符在它们交叉后被打印。在上部，str[i]打印在 str[j]之前，在下部，str[j]打印在 str[i]之前。 以下是以上思路的实现。
c
// c program to print cross pattern
#include
using namespace std;
// function to print given string in cross pattern
// length of string must be odd
void printpattern(string str)
{
int len = str.length();
// i goes from 0 to len and j goes from len-1 to 0
for (int i=0,j=len-1; i<=len,j>=0; i ,j--)
{
// to print the upper part. this loop runs
// till middle point of string (i and j become
// same
if (ij)
{
// print j spaces
for (int x = j-1; x>=0; x--)
cout << " ";
// print j'th character
cout << str[j];
// print i-j-1 spaces
for (int x=0; x
java 语言(一种计算机语言，尤用于创建网站)
// java program to
// print cross pattern
class gfg
{
// function to print given
// string in cross pattern
static void pattern(string str,
int len)
{
// i and j are the indexes
// of characters to be
// displayed in the ith
// iteration i = 0 initially
// and go upto length of string
// j = length of string initially
// in each iteration of i,
// we increment i and decrement j,
// we print character only
// of k==i or k==j
for (int i = 0; i < len; i )
{
int j = len - 1 - i;
for (int k = 0; k < len; k )
{
if (k == i || k == j)
system.out.print(str.charat(k));
else
system.out.print(" ");
}
system.out.println("");
}
}
// driver code
public static void main (string[] args)
{
string str = "geeksforgeeks";
int len = str.length();
pattern(str, len);
}
}
// this code is contributed
// by smitha
python 3
# python 3 program to
# print cross pattern
# function to print given
# string in cross pattern
def pattern(str, len):
# i and j are the indexes
# of characters to be
# displayed in the ith
# iteration i = 0 initially
# and go upto length of string
# j = length of string initially
# in each iteration of i, we
# increment i and decrement j,
# we print character only of
# k==i or k==j
for i in range(0, len):
j = len -1 - i
for k in range(0, len):
if (k == i or k == j):
print(str[k],
end = "")
else:
print(end = " ")
print(" ")
# driver code
str = "geeksforgeeks"
len = len(str)
pattern(str, len)
# this code is contributed
# by smitha
c
// c# program to print
// cross pattern
using system;
class gfg
{
// function to print given
// string in cross pattern
static void pattern(string str,
int len)
{
// i and j are the indexes
// of characters to be
// displayed in the ith
// iteration i = 0 initially
// and go upto length of string
// j = length of string initially
// in each iteration of i, we
// increment i and decrement j,
// we print character only of
// k==i or k==j
for (int i = 0; i < len; i )
{
int j = len - 1 - i;
for (int k = 0; k < len; k )
{
if (k == i || k == j)
console.write(str[k]);
else
console.write(" ");
}
console.write("\n");
}
}
// driver code
public static void main ()
{
string str = "geeksforgeeks";
int len = str.length;
pattern(str, len);
}
}
// this code is contributed by smitha
服务器端编程语言（professional hypertext preprocessor 的缩写）
= 0;
$i , $j--)
{
// to print the upper part.
// this loop runs till middle point
// of string i and j become same
if ($i < $j)
{
// print i spaces
for ($x = 0; $x < $i; $x )
echo " ";
// print i'th character
echo $str[$i];
// print j-i-1 spaces
for ( $x = 0; $x < $j - $i - 1;
$x )
echo " ";
// print j'th character
echo $str[$j]."\n";
}
// to print center point
if ($i == $j)
{
// print i spaces
for ($x = 0; $x < $i; $x )
echo " ";
// print middle character
echo $str[$i]."\n";
}
// to print lower part
else if ($i > $j)
{
// print j spaces
for ($x = $j - 1; $x >= 0;
$x--)
echo " ";
// print j'th character
echo $str[$j];
// print i-j-1 spaces
for ( $x = 0; $x < $i - $j - 1;
$x )
echo " ";
// print i'h character
echo $str[$i]."\n";
}
}
}
// driver code
printpattern("geeksforgeeks");
// this code is contributed by mits
?>
java 描述语言
输出:
g s
e k
e e
k e
s g
f r
o
f r
s g
k e
e e
e k
g s
备选方案:
c
// cpp program to print cross pattern
#include
using namespace std;
// function to print given string in
// cross pattern
void pattern(string str, int len){
// i and j are the indexes of characters
// to be displayed in the ith iteration
// i = 0 initially and go upto length of
// string
// j = length of string initially
// in each iteration of i, we increment
// i and decrement j, we print character
// only of k==i or k==j
for (int i = 0; i < len; i )
{
int j = len -1 - i;
for (int k = 0; k < len; k )
{
if (k == i || k == j)
cout << str[k];
else
cout << " ";
}
cout << endl;
}
}
// driver code
int main ()
{
string str = "geeksforgeeks";
int len = str.size();
pattern(str, len);
return 0;
}
// this code is contributed by satinder kaur
java 语言(一种计算机语言，尤用于创建网站)
// java program to print cross pattern
class gfg
{
// function to print given
// string in cross pattern
static void pattern(string str, int len)
{
// i and j are the indexes of
// characters to be displayed
// in the ith iteration i = 0
// initially and go upto length
// of string j = length of string
// initially in each iteration
// of i, we increment i and decrement
// j, we print character only
// of k==i or k==j
for (int i = 0; i < len; i )
{
int j = len -1 - i;
for (int k = 0; k < len; k )
{
if (k == i || k == j)
system.out.print(str.charat(k));
else
system.out.print(" ");
}
system.out.println("");
}
}
// driver code
public static void main(string[] args)
{
string str = "geeksforgeeks";
int len = str.length();
pattern(str, len);
}
}
// this code is contributed by 29ajaykumar
python 3
# python 3 program to print cross pattern
# function to print given string in
# cross pattern
def pattern(st, length):
# i and j are the indexes of characters
# to be displayed in the ith iteration
# i = 0 initially and go upto length of
# string
# j = length of string initially
# in each iteration of i, we increment
# i and decrement j, we print character
# only of k==i or k==j
for i in range(length):
j = length -1 - i
for k in range(length):
if (k == i or k == j):
print(st[k],end="")
else:
print(" ",end="")
print()
# driver code
if __name__ == "__main__":
st = "geeksforgeeks"
length = len(st)
pattern(st, length)
c
// c# program to print cross pattern
using system;
class gfg
{
// function to print given
// string in cross pattern
static void pattern(string str, int len)
{
// i and j are the indexes of
// characters to be displayed
// in the ith iteration i = 0
// initially and go upto length
// of string j = length of string
// initially in each iteration
// of i, we increment i and decrement
// j, we print character only
// of k==i or k==j
for (int i = 0; i < len; i )
{
int j = len -1 - i;
for (int k = 0; k < len; k )
{
if (k == i || k == j)
console.write(str[k]);
else
console.write(" ");
}
console.writeline("");
}
}
// driver code
public static void main(string[] args)
{
string str = "geeksforgeeks";
int len = str.length;
pattern(str, len);
}
}
// this code is contributed by rajput-ji
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
g s
e k
e e
k e
s g
f r
o
f r
s g
k e
e e
e k
g s
pg电子试玩链接的解决方案 3 :这个问题也可以通过观察字符是沿着左右对角线打印的，只要我们把图案封装在一个矩阵内就可以解决。现在，如果字符串的长度是 len ，那么该模式可以包含在有序的方形矩阵 len 中。
- 左对角线上的元素可以通过条件 ( i==j ) 来访问，其中 i 和 j 分别是行号和列号。
- 右对角线上的元素可以通过条件 (i j == len-1) 访问。
因此，运行顺序为 len 的嵌套循环，用各自的字符填充满足上述两个条件的位置，并用空格填充其余位置。
以下是上述办法的实施情况:
卡片打印处理机（card print processor 的缩写）
// c program to print the given pattern
#include
using namespace std;
// function to print the given
// string in respective pattern
void printpattern(string str, int len)
{
for(int i = 0; i < len; i )
{
for(int j = 0; j < len; j )
{
// print characters at corresponding
// places satisfying the two conditions
if((i == j) || (i j == len-1))
cout<
java 语言(一种计算机语言，尤用于创建网站)
// java program to print the given pattern
import java.io.*;
class gfg
{
// function to print the given
// string in respective pattern
static void printpattern(string str, int len)
{
for(int i = 0; i < len; i )
{
for(int j = 0; j < len; j )
{
// print characters at corresponding
// places satisfying the two conditions
if((i == j) || (i j == len - 1))
system.out.print(str.charat(j));
// print blank space at rest of places
else
system.out.print(" ");
}
system.out.println();
}
}
// driver code
public static void main (string[] args)
{
string str = "geeksforgeeks";
int len = str.length();
printpattern(str, len);
}
}
// this code is contributed by rag2127.
python 3
# python3 program to print the given pattern
# function to print the given
# string in respective pattern
def printpattern (str, len) :
for i in range(len) :
for j in range(len) :
# print characters at corresponding
# places satisfying the two conditions
if ((i == j) or (i j == len - 1)) :
print(str[j], end = "")
# print blank space at rest of places
else :
print(" ", end = "")
print()
str = "geeksforgeeks"
len = len(str)
printpattern(str, len)
# this code is contributed by divyeshrabadiya07.
c
// c# program to print the given pattern
using system;
public class gfg
{
// function to print the given
// string in respective pattern
static void printpattern(string str, int len)
{
for(int i = 0; i < len; i )
{
for(int j = 0; j < len; j )
{
// print characters at corresponding
// places satisfying the two conditions
if((i == j) || (i j == len - 1))
console.write(str[j]);
// print blank space at rest of places
else
console.write(" ");
}
console.writeline();
}
}
// driver code
static public void main ()
{
string str = "geeksforgeeks";
int len = str.length;
printpattern(str, len);
}
}
// this code is contributed by avanitrachhadiya2155
java 描述语言
输出:
g s
e k
e e
k e
s g
f r
o
f r
s g
k e
e e
e k
g s
https://youtu . be/7 gyjmrezos
本文由 供稿。如果你发现任何不正确的地方，请写评论，或者你想分享更多关于上面讨论的话题的信息
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
