要求:通过循环输出一下图形
可以看成三个三角形的组合
#!/bin/bash
for i in {1..9}
do
for j in {9..1}
do
[ $j -gt $i ] && echo -n " "
done
for m in {1..9}
do
[ $m -le $i ] && echo -n "*"
done
for k in {1..9}
do
[ $k -lt $i ] && echo -n "*"
done
echo
done
结果如下: