微軟2014實習生及秋令營技術類職位線上測試--String reorder

題目1 : String reorder

時間限制:10000ms單點時限:1000ms記憶體限制:256MB

Description

For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’).

Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met,
1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order).
2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment.

Your program should output string “<invalid input string>” when the input contains any invalid characters (i.e., outside the '0'-'9' and 'a'-'z' range).

Input

Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters.

Output

For each case, print exactly one line with the reordered string based on the criteria above.

範例輸入

aabbccdd007799aabbccddeeff113355zz1234.89898abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

範例輸出

abcdabcd013579abcdefz013579abcdefz<invalid input string>abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

解題思路：

關鍵是對字串的熟練操作，若是自己不熟悉，一遍測試一遍寫，就慢很多。

演算法中用到的重要操作細節舉例：

string str="abcdefg";

string::iterator iter=str.begin();

str.erase(iter); //執行這條語句，會刪除字元 'a'， 返回的指向下一個字元'b'的迭代器

cout<<*iter<<endl; //特別注意這裡iter會指向字元  'b' 了

下面附上自己實現的一個代碼，寫的不好，望見諒

#include <iostream>#include <cstdio> //包含語言重新導向函數freopen的庫#include <vector>#include <algorithm>#include <string>#include <cstring>#include <iterator>using namespace std;bool valiade(char a){if((a>='a' && a<='z') ||(a>='0' && a<='9')){return true;}else{return false;}}int main(){freopen("input.txt","r",stdin); //重新導向輸入資料流//freopen("ouput.txt","w",stdout);string str;while(cin>>str){bool flag=true;sort(str.begin(),str.end());//cout<<str<<endl;string temp;while(!str.empty()){string::iterator iter=str.begin();char lastch=*(iter);if(!valiade(lastch)){flag=false;break;}temp += *iter;(str.erase(iter));for(string::iterator it=str.begin(); it < str.end();){ if(!valiade(lastch)){flag=false;goto here;}if( *it != lastch ){temp += *it;lastch = *it;str.erase(it);}else{++it;}}}here: if(flag){cout<<temp<<endl;}else{cout<<"<invalid input string>"<<endl;}}return 0; }

自己又修改了一下，使代碼更好看了些。 ^_^

#include <iostream>#include <cstdio> //包含語言重新導向函數freopen的庫#include <algorithm>#include <string>#include <iterator>using namespace std;bool valiade(char a){if((a>='a' && a<='z') ||(a>='0' && a<='9')){return true;}else{return false;}}int main(){freopen("input.txt","r",stdin); //重新導向輸入資料流//freopen("ouput.txt","w",stdout);string str;while(cin>>str){bool flag=true;sort(str.begin(),str.end());string temp;while(!str.empty()){string::iterator iter=str.begin();char lastch=' ';do{if( *iter != lastch){temp += *iter;lastch = *iter;str.erase(iter);}else{iter++;}if(!valiade(lastch)){// validate laterflag=false;goto done;}}while(iter<str.end());}done: if(flag){cout<<temp<<endl;}else{cout<<"<invalid input string>"<<endl;}}return 0; }

若你對此問題有疑問，或寫出了更緊湊更好的代碼，歡迎交流。轉載請註明出處。