如何使排列在bash与N！输入？ [英] How to make permutation in bash with N! input?

问题描述

我必须做在bash置换以EVAL和序列命令。所以，我不得不做出第一个可能包含相同的号码置换的话，我必须以某种方式进行过滤。

I have to make a permutation in bash with "eval" and "seq" commands. So I have to make first a permutation that could contain the same numbers, then I have to filter it somehow.

老师告诉我，我必须使用两个周期/周期，一个在其他。但我不知道该怎么办了。

The teacher told me I have to use two cycle/period, one in the other. But I don't know how to do it.

输入是这样的：

3

输出应该是：

123
132
213
231
312
321

如果有人可以帮助我，这将是伟大的！

If someone could help me, that would be great!

编辑：
我想知道如何使用这些命令做到这一点，我的朋友去到我这个样子： http://pastebin.com/ wM2U1SuT
老师告诉我它太好了，不是在这个层面上..我应该与序列和eval做到这一点，即使它不是那么好。第二个问题是，输入为123，而不是3，如3！..

I want to know how to do it with these commands, my friend made it to me like this: http://pastebin.com/wM2U1SuT the teacher told me its too good, were not on this level.. i should do it with seq and eval, even if its not that good then. The second problem was that the input was 123, not 3, like 3!..

推荐答案

如果某人的好奇，继承人的解决方案：

If someones curious, heres the solution:

input=3
for i in $(eval echo " {1..$input}{1..$input}{1..$input} "); do
    OK=yes
    for pos1 in $(seq 0 $((x,1)) ); do
        for pos2 in $(seq 0 $((x,1)) ); do
            if [ $pos1 != $pos2 ]; then
                if [ ${i:$pos1:1} == ${i:$pos2:1} ]; then
                    OK=no
                fi
            fi
        done
    done
    if [ $OK = yes ]; then
        echo $i
    fi

done

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