有没有办法让代码运行得更快？我已经尝试了一切！
基本上我想做的是： 计算52个变量的所有组合，但仅显示所有数字仅列出一次的组合！所以不可能是fx：两个1或者三个49＆＃39>
#!/bin/bash
#Enter here how many cards you use
howmanycards=52;
let run=$howmanycards+1;
i=1;
u=52;
totalrun=0;
SECONDS=0;
while [ $i -lt $run ]; do
let card$i=$u;
let i++;
let u--;
done
while [ -lt ]; do
let i=1;
if [ "$card1" -gt "52" ]; then let card2++;let card1=1; fi
if [ "$card2" -gt "52" ]; then let card3++;let card2=1; fi
if [ "$card3" -gt "52" ]; then let card4++;let card3=1; fi
if [ "$card4" -gt "52" ]; then let card5++;let card4=1; fi
if [ "$card5" -gt "52" ]; then let card6++;let card5=1; fi
if [ "$card6" -gt "52" ]; then let card7++;let card6=1; fi
if [ "$card7" -gt "52" ]; then let card8++;let card7=1; fi
if [ "$card8" -gt "52" ]; then let card9++;let card8=1; fi
if [ "$card9" -gt "52" ]; then let card10++;let card9=1; fi
if [ "$card10" -gt "52" ]; then let card11++;let card10=1; fi
if [ "$card11" -gt "52" ]; then let card12++;let card11=1; fi
if [ "$card12" -gt "52" ]; then let card13++;let card12=1; fi
if [ "$card13" -gt "52" ]; then let card14++;let card13=1; fi
if [ "$card14" -gt "52" ]; then let card15++;let card14=1; fi
if [ "$card15" -gt "52" ]; then let card16++;let card15=1; fi
if [ "$card16" -gt "52" ]; then let card17++;let card16=1; fi
if [ "$card17" -gt "52" ]; then let card18++;let card17=1; fi
if [ "$card18" -gt "52" ]; then let card19++;let card18=1; fi
if [ "$card19" -gt "52" ]; then let card20++;let card19=1; fi
if [ "$card20" -gt "52" ]; then let card21++;let card20=1; fi
if [ "$card21" -gt "52" ]; then let card22++;let card21=1; fi
if [ "$card22" -gt "52" ]; then let card23++;let card22=1; fi
if [ "$card23" -gt "52" ]; then let card24++;let card23=1; fi
if [ "$card24" -gt "52" ]; then let card25++;let card24=1; fi
if [ "$card25" -gt "52" ]; then let card26++;let card25=1; fi
if [ "$card26" -gt "52" ]; then let card27++;let card26=1; fi
if [ "$card27" -gt "52" ]; then let card28++;let card27=1; fi
if [ "$card28" -gt "52" ]; then let card29++;let card28=1; fi
if [ "$card30" -gt "52" ]; then let card31++;let card30=1; fi
if [ "$card31" -gt "52" ]; then let card32++;let card31=1; fi
if [ "$card32" -gt "52" ]; then let card33++;let card32=1; fi
if [ "$card33" -gt "52" ]; then let card34++;let card33=1; fi
if [ "$card34" -gt "52" ]; then let card35++;let card34=1; fi
if [ "$card35" -gt "52" ]; then let card36++;let card35=1; fi
if [ "$card36" -gt "52" ]; then let card37++;let card36=1; fi
if [ "$card37" -gt "52" ]; then let card38++;let card37=1; fi
if [ "$card38" -gt "52" ]; then let card39++;let card38=1; fi
if [ "$card39" -gt "52" ]; then let card40++;let card39=1; fi
if [ "$card40" -gt "52" ]; then let card41++;let card40=1; fi
if [ "$card41" -gt "52" ]; then let card42++;let card41=1; fi
if [ "$card42" -gt "52" ]; then let card43++;let card42=1; fi
if [ "$card43" -gt "52" ]; then let card44++;let card43=1; fi
if [ "$card44" -gt "52" ]; then let card45++;let card44=1; fi
if [ "$card45" -gt "52" ]; then let card46++;let card45=1; fi
if [ "$card46" -gt "52" ]; then let card47++;let card46=1; fi
if [ "$card47" -gt "52" ]; then let card48++;let card47=1; fi
if [ "$card48" -gt "52" ]; then let card49++;let card48=1; fi
if [ "$card49" -gt "52" ]; then let card50++;let card49=1; fi
if [ "$card50" -gt "52" ]; then let card51++;let card50=1; fi
if [ "$card51" -gt "52" ]; then let card52++;let card51=1; fi
while [ $i -lt $run ]; do
temp=${card}${i};
if [ "$temp" = "$card1" ] || [ "$temp" = "$card2" ] ||
[ "$temp" = "$card3" ] || [ "$temp" = "$card4" ] ||
[ "$temp" = "$card5" ] || [ "$temp" = "$card6" ] ||
[ "$temp" = "$card7" ] || [ "$temp" = "$card8" ] ||
[ "$temp" = "$card9" ] || [ "$temp" = "$card10" ] ||
[ "$temp" = "$card11" ] || [ "$temp" = "$card12" ] ||
[ "$temp" = "$card13" ] || [ "$temp" = "$card14" ] ||
[ "$temp" = "$card15" ] || [ "$temp" = "$card16" ] ||
[ "$temp" = "$card17" ] || [ "$temp" = "$card18" ] ||
[ "$temp" = "$card19" ] || [ "$temp" = "$card20" ] ||
[ "$temp" = "$card21" ] || [ "$temp" = "$card22" ] ||
[ "$temp" = "$card23" ] || [ "$temp" = "$card24" ] ||
[ "$temp" = "$card25" ] || [ "$temp" = "$card26" ] ||
[ "$temp" = "$card27" ] || [ "$temp" = "$card28" ] ||
[ "$temp" = "$card29" ] || [ "$temp" = "$card30" ] ||
[ "$temp" = "$card31" ] || [ "$temp" = "$card32" ] ||
[ "$temp" = "$card33" ] || [ "$temp" = "$card34" ] ||
[ "$temp" = "$card35" ] || [ "$temp" = "$card36" ] ||
[ "$temp" = "$card37" ] || [ "$temp" = "$card38" ] ||
[ "$temp" = "$card39" ] || [ "$temp" = "$card40" ] ||
[ "$temp" = "$card41" ] || [ "$temp" = "$card42" ] ||
[ "$temp" = "$card43" ] || [ "$temp" = "$card44" ] ||
[ "$temp" = "$card45" ] || [ "$temp" = "$card46" ] ||
[ "$temp" = "$card47" ] || [ "$temp" = "$card48" ] ||
[ "$temp" = "$card49" ] || [ "$temp" = "$card50" ] ||
[ "$temp" = "$card51" ] || [ "$temp" = "$card52" ]; then
let usefull++;
fi
let i++;
done
if [ $usefull -gt 51 ]; then
echo "[loops($totalrun)] $card52-$card51-$card50-$card49-$card48-$card47-$card46-$card45-$card44-$card43-$card42-$card41-$card40-$card39-$card38-$card37-$card36-$card35-$card34-$card33-$card32-$card31-$card30-$card29-$card28-$card27-$card26-$card25-$card24-$card23-$card22-$card21-$card20-$card19-$card18-$card17-$card16-$card15-$card14-$card13-$card12-$card11-$card10-$card9-$card8-$card7-$card6-$card5-$card4-$card3-$card2-$card1";
fi
let usefull=0;
if [ "$card52" -gt "52" ]; then
echo " ";
duration=$SECONDS
echo "$(($duration / 60)) min and $(($duration % 60)) sec";
exit;
fi
let card1++;
let totalrun++;
done
我曾尝试创建代码优先级，但它似乎无法发挥作用！
但我得知的是，进行检查的代码，如果有多个数字，则使用了大量的性能！我不知道该怎么做！
感谢您的帮助。
答案 0 :(得分：0)
首先，我认为你需要一个可行的实现来获得bash中的组合。这是一个使用关联数组的方法。我实现它来模仿this Java code：
#!/bin/bash
function combine() {
local index="$1"
local -a prefix=("${!2}")
local -a postfix=("${!3}")
local i=$index
while [ $i -lt ${#prefix[@]} ]; do
postfix+=(${prefix[@]:$i:1})
echo ${postfix[@]}
combine "$((i + 1))" prefix[@] postfix[@]
local postfix_last_char="$((${#postfix[@]} - 1))"
postfix=(${postfix[@]:0:$postfix_last_char})
((i++))
done
}
function get_combinations() {
local -a prefix_table=("${!1}")
local -a postfix_table=()
local index=0
combine 0 prefix_table[@] postfix_table[@]
}
function example_use() {
# Get all combinations of the numbers 0 to 9.
foo=({0..9})
get_combinations foo[@]
}
现在，人们可以开始考虑修改combine()函数以强加一个额外的条件，我们只打印出集合（只包含不同元素的数组）。所以我们可以替换这行：
echo ${postfix[@]}
通过新电话：
echo_if_is_a_set postfix[@]
实现方式如下：
function echo_if_is_a_set() {
local -a input=("${!1}")
local -a entry_count_lookup
# Tally up the occurrences of each item.
for item in ${input[@]}; do
local current_count=${entry_count_lookup[$item]:-0}
((current_count++))
entry_count_lookup[$item]=$current_count
done
for value in ${!entry_count_lookup[@]}; do
if [ ${entry_count_lookup[$value]} -ne 1 ]; then
return
fi
done
echo ${input[@]}
}
好的，但还没完成。如果在重复元素中传递，我们可能仍然会打印两次相同的组合。例如，如果您这样做：
foo=(1 1 1)
get_combinations foo[@]
所以，我们只需调用uniq，将get_combinations更改为：
function get_combinations() {
local -a prefix_table=("${!1}")
local -a postfix_table=()
local index=0
combine 0 prefix_table[@] postfix_table[@] | sort | uniq
}
而且，我们已经完成了。使用输入({1..3})运行此打印：
1
1 2
1 2 3
1 3
2
2 3
3
但是这个实现的优点在于因为它使用了关联数组，所以我们实际上可以是通用的，并输入类似(2 Racecar 34)的东西，它会返回：
2
2 34
2 Racecar
2 Racecar 34
34
Racecar
Racecar 34
无论如何，作为最后的评论，我会说从这样的事情开始，然后你可以优化算法以提高效率。
另外，Bash可能并不适合这种语言。 :-D