Python Programming Questions-stitching the smallest dictionary order

Source: Niu Ke Net

For a given array of strings, find a concatenation order so that all small strings are stitched into a large string that is the smallest of all possible stitching in the dictionary order.

Given a string array STRs, given its size, return the concatenation of the strings.

Test examples:

["abc", "De"],2

"ABCDE"

Python

Code

#-*-Coding:utf-8-*-class prior:def findsmallest (self, STRs, n): # write code here RESSTR = ""                    For I in Xrange (1,len (STRs)): for J in Xrange (0,i): if (Strs[i]+strs[j]) < (Strs[j]+strs[i]): STRS[I],STRS[J] = Strs[j],strs[i] return '. Join (STRs)

Shell implementations

Code

#/bin/bash#by xianwei#2017-9-4cat <<eof topic for a given array of strings, find a concatenation order so that all small strings are stitched into a large string that is the smallest of all possible stitching in the dictionary order. Given a string array strs, given its size, return the concatenation of the strings. Test sample: ["abc", "De"],2 "ABCDE" eofa= ("Kid" "yqt" "" I "" K ") for ((i=0;i<${#a [*]};i++)] do if [[" j=0;j<i;j++]    a[$i]}${a[$j]} "<" ${a[$j]}${a[$i]} "] then tmp=${a[$i]}echo $tmp a[$i]=${a[$j]} a[$j]= $tmp fi Donedoneecho ${a[*]}

C + + implementation

Code

Class prior {public:    string findsmallest (Vector<string> strs,  int n)  {        // write code here         string A;         for  (Int i = 0; i < strs.size (); i++)          {            for  (int j  = i; j < strs.size ();  j++)              {                 if  ((Strs[i] + strs[j]) > (strs[j] + strs[i])//If k+kid >  Kid +k, Exchange                       swap (strs[i],strs[j]);            }         }        for  (int i  = 0; i < strs.size ();  i++)         {             A = A + strs[i];         }        return a ;    }};

This article is from "operations to Development" blog, please be sure to keep this source http://237085.blog.51cto.com/227085/1962689

Python Programming Questions-stitching the smallest dictionary order