微软2014实习生及秋令营技术类职位在线测试 原创解答 题目1 : String reorder

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解题思路：利用vector，自定义类，建立一个统计所有字符map，将字符存入，并将字符按ascii码值排序。输出时按照字符顺序输出，输出一个字符要将相应的对应统计数自减1，直到所有字符输出完毕。

时间限制:10000ms

单点时限:1000ms

内存限制: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.

样例输入

aabbccdd007799aabbccddeeff113355zz1234.89898abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

样例输出

abcdabcd013579abcdefz013579abcdefz<invalid input string>abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

解答如下：

#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>#include <cstring>using namespace std;bool parseChar(char * c);class CInputStr{public:char m_cStr;int m_iNum;};struct Cmp_by_name   //为排序编写的结构体{bool operator()(const CInputStr &a, const CInputStr &b){return a.m_cStr<b.m_cStr;}};int main() {char c[5][100];  char * result[5];int inumcount;  int index=0;  int temp;  int minnum=10000;for (int i=0;i<5;i++){cin.getline(c[i],100);if (parseChar(c[i])){}else{char b[100]="<invalid input string>";strcpy(c[i],b);}}for (int i=0;i<5;i++){cout<<c[i]<<endl;}}bool parseChar(char * c){int icount=0;char * result=c;vector <CInputStr> vInput;while(c[icount]!='\0'){if ((c[icount]>='0'&&c[icount]<='9')||(c[icount]>='a'&&c[icount]<='z')){if (0==vInput.size()){CInputStr cinput;cinput.m_cStr=c[icount];cinput.m_iNum=1;vInput.push_back(cinput);}else{for (int i=0;i<vInput.size();i++){if (c[icount]==vInput[i].m_cStr){vInput[i].m_iNum++;break;}if (i==vInput.size()-1){CInputStr cinput;cinput.m_cStr=c[icount];cinput.m_iNum=1;vInput.push_back(cinput);break;}}}}else{//cout<<"<invalid input string>";return false;}icount++;}int iVectorCount=0;icount=0;sort(vInput.begin(),vInput.end(),Cmp_by_name());while (0!=vInput.size()){if (0!=vInput[iVectorCount].m_iNum){result[icount]=vInput[iVectorCount].m_cStr;vInput[iVectorCount].m_iNum-=1;icount++;}else{if(0==vInput[vInput.size()-1].m_iNum){vInput.pop_back();}}if (iVectorCount+1>=vInput.size()){iVectorCount=0;}else{iVectorCount++;}}vInput.empty();return true;}