一道Topcoder Algorithms 550分的题

本文介绍了一个编程问题,即如何通过填充缺失字符使两个字符串成为字谜。任务要求在多个可能的解决方案中选择字母顺序最优的一组结果。

(我自己实现的代码在另外一篇文章) 

Problem Statement
    
Two words are said to be anagrams if one word can be formed by rearranging the letters of the other word. For example, "AABC" and "CABA" are anagrams.
You will be given two strings, word1 and word2, representing two words. These two strings will contain only uppercase letters ('A'-'Z') and '.' characters which denote missing letters. Replace the '.' characters with uppercase letters such that the two words become anagrams and return the result as a vector <string> containing exactly two elements. The first element should correspond to word1 and the second element should correspond to word2. If there are multiple possible return values, choose the one in which the first element comes first alphabetically. If ties still exist, choose the one among them in which the second element comes first alphabetically. If it is impossible to make the two words into anagrams, return an empty vector <string>. Note that two equal words are considered anagrams.
Definition
    
Class:                                   AnagramCompletion
Method:                                complete
Parameters:                        string, string
Returns:                               vector <string>
Method signature:              vector <string> complete(string word1, string word2)
(be sure your method is public)
    

Constraints
-
word1 will contain between 1 and 50 characters, inclusive.
-
word2 will contain the same number of characters as word1.
-
word1 and word2 will contain only uppercase letters ('A'-'Z') and '.' characters.
-
There will be at least one '.' in at least one of the two words.


Examples
0)
"AB.AC."
"ABD..E"
Returns: {"ABDACE", "ABDACE" }
The letters 'D' and 'E' appear in the second word but not in the first one, so we place them in the two empty spots. Similarly the letter 'C' must be present in the second word. The letter 'A' must appear a second time in the second word. Ties are broken alphabetically.


1)
"ABC..."
"DEF..."
Returns: {"ABCDEF", "DEFABC" }

2)
"......"
"......"
Returns: {"AAAAAA", "AAAAAA" }
All the letters are missing, so we can replace them all with the letter 'A'.


3) 
"TOPCODER"
"D.E..TR."
Returns: {"TOPCODER", "DCEOOTRP" }

4)
"ABC."
"DEF."
Returns: { }
Each word is missing three letters from the other word, but there is only one empty spot in each of them.


5)
"TEFAT..PVSKKJT.QBJEB..NPN..NBL"
"...B...E.ND.LNE...HW.ANTB.TKBD"
Returns: {"TEFATAAPVSKKJTAQBJEBDDNPNHWNBL", "AAABFJJEKNDPLNEPQSHWTANTBVTKBD" }

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

这个是完整源码 python实现 Flask,Vue 【python毕业设计】基于Python的Flask+Vue物业管理系统 源码+论文+sql脚本 完整版 数据库是mysql 本文首先实现了基于Python的Flask+Vue物业管理系统技术的发展随后依照传统的软件开发流程,最先为系统挑选适用的言语和软件开发平台,依据需求析开展控制模块制做和数据库查询构造设计,随后依据系统整体功能模块的设计,制作系统的功能模块图、E-R图。随后,设计框架,依据设计的框架撰写编码,完成系统的每个功能模块。最终,对基本系统开展了检测,包含软件性能测试、单元测试和性能指标。测试结果表明,该系统能够实现所需的功能,运行状况尚可并无明显缺点。本文首先实现了基于Python的Flask+Vue物业管理系统技术的发展随后依照传统的软件开发流程,最先为系统挑选适用的言语和软件开发平台,依据需求析开展控制模块制做和数据库查询构造设计,随后依据系统整体功能模块的设计,制作系统的功能模块图、E-R图。随后,设计框架,依据设计的框架撰写编码,完成系统的每个功能模块。最终,对基本系统开展了检测,包含软件性能测试、单元测试和性能指标。测试结果表明,该系统能够实现所需的功能,运行状况尚可并无明显缺点。本文首先实现了基于Python的Flask+Vue物业管理系统技术的发展随后依照传统的软件开发流程,最先为系统挑选适用的言语和软件开发平台,依据需求析开展控制模块制做和数据库查询构造设计,随后依据系统整体功能模块的设计,制作系统的功能模块图、E-R图。随后,设计框架,依据设计的框架撰写编码,完成系统的每个功能模块。最终,对基本系统开展了检测,包含软件性能测试、单元测试和性能指标。测试结果表明,该系统能够实现所需的功能,运行状况尚可并无明显缺点。本文首先实现了基于Python的Flask+Vue物业管理系统技术的发
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