bash 检查字符串是否为回文
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checking if a string is a palindrome
提问by Little Child
I am trying to check if a string is a palindrome in bash. Here is what I came up with:
我正在尝试检查字符串是否是 bash 中的回文。这是我想出的：
#!/bin/bash
read -p "Enter a string: " string
if [[ $string|rev == $string ]]; then
echo "Palindrome"
fi
Now, echo $string|revgives reversed string. My logic was to use it in the condition for if. That did not work out so well.
现在，echo $string|rev给出反向字符串。我的逻辑是在 for 的条件下使用它if。效果不是很好。
So, how can I store the "returned value" from revinto a variable? or use it directly in a condition?
那么，如何将“返回值”存储rev到变量中？还是直接在条件下使用？
回答by glenn Hymanman
A bash-only implementation:
仅 bash 的实现：
is_palindrome () {
local word=
local len=$((${#word} - 1))
local i
for ((i=0; i <= (len/2); i++)); do
[[ ${word:i:1} == ${word:len-i:1} ]] || return 1
done
return 0
}
for word in hello kayak; do
if is_palindrome $word; then
echo $word is a palindrome
else
echo $word is NOT a palindrome
fi
done
Inspired by gniourf_gniourf:
受 gniourf_gniourf 的启发：
is_palindrome() {
(( ${#1} <= 1 )) && return 0
[[ ${1:0:1} != ${1: -1} ]] && return 1
is_palindrome ${1:1: 1}
}
I bet the performance of this truly recursive call really sucks.
我敢打赌这个真正递归调用的性能真的很糟糕。
回答by janos
Another variation without echoand unnecessary quoting within [[ ... ]]:
另一种没有echo和不必要的引用的变体[[ ... ]]：
#!/bin/bash
read -p "Enter a string: " string
if [[ $(rev <<< "$string") == "$string" ]]; then
echo Palindrome
fi
回答by that other guy
Use $(command substitution):
使用$(command substitution)：
#!/bin/bash
read -p "Enter a string: " string
if [[ "$(echo "$string" | rev)" == "$string" ]]; then
echo "Palindrome"
fi
回答by Alex
Maybe it is not the best implementation, but if you need something with sh only
也许这不是最好的实现，但如果你只需要 sh 的东西
#!/bin/sh
#get character <str> <num_of_char>. Please, remember that indexing is from 1
get_character() {
echo "" | cut -c ""
}
for i in $(seq $((${#1} / 2))); do
if [ "$(get_character "" "$i")" != "$(get_character "" $((${#1} - i + 1)))" ]; then
echo "NO"
exit 0
fi
done
echo "YES"
and canonical way with bash as well
以及 bash 的规范方式
for i in $(seq 0 $((${#1} / 2 - 1))); do
if [ "${1:$i:1}" != "${1:$((${#1} - i - 1)):1}" ]; then
echo "NO"
exit 0
fi
done
echo "YES"
