微软2014年4月 实习生招聘机试题 1.String reorder

《1.String reorder》

Time Limit: 10000ms
Case Time Limit: 1000ms
Memory Limit: 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.

 

Sample In

aabbccdd
007799aabbccddeeff113355zz
1234.89898
abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

Sample Out

abcdabcd
013579abcdefz013579abcdefz
<invalid input string>
abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

 

个人给出的一种解：

#include <fstream>#include <iostream>#include <string>using namespace std;#define USE_IFSTREAM class CHAR_COUNTER{public:char charValue;long charCount;};//计数：这里我们需要降序int compare_CHAR_COUNTER(const void * ptr1,const void * ptr2){long value1=(*(CHAR_COUNTER*)ptr1).charCount;long value2=(*(CHAR_COUNTER*)ptr2).charCount;return (value1>value2)?-1:(value1==value2?0:1);}//字符：这里排成降序，后面颠倒顺序后(按升序)输出int compare_char(const void * ptr1,const void * ptr2){char value1 = *(char*)ptr1;char value2=  *(char*)ptr2;return (value1>value2)?-1:(value1==value2?0:1);}void main(){CHAR_COUNTER charArray[36];char charSet[36+1];int valueCharacters=0;string str ;long length = 0;long i = 0;long j = 0;int  k = 0;int w;long curOutPutTime = 0;long maxOutPutTime =0;long pausePoint =0;#ifdef USE_IFSTREAMifstream ifs=ifstream("testCase.txt");#endifwhile(1){i=0;valueCharacters = 0;memset(charArray,0,sizeof(charArray));for( i=0;i<36;i++){if(i<10)charArray[i].charValue = char('0'+ i);else charArray[i].charValue = char('a'+(i-10));charArray[i].charCount = 0;}#ifndef USE_IFSTREAMgetline(cin,str);#else  if(ifs.eof())break;getline(ifs,str);#endiflength = str.length();i=0;while(i<length){char charTmp = str[i];if(charTmp>='0' && charTmp<='9')charArray[charTmp-'0'].charCount++;else if(charTmp>='a' && charTmp<='z')charArray[charTmp-'a'+10].charCount++;else{ cout<<"<invalid input string>"<<endl;goto again;}i++;}qsort(charArray,sizeof(charArray)/sizeof(CHAR_COUNTER),sizeof(CHAR_COUNTER),compare_CHAR_COUNTER);memset(charSet,0,sizeof(charSet));i=0;while(charArray[i].charCount){charSet[i]=charArray[i].charValue;i++;}valueCharacters = i;//最少 1qsort(charSet,sizeof(charSet)/sizeof(char)-1,sizeof(char),compare_char);//最后一个不参与排序for(i=0;i<(valueCharacters/2);i++){ //逆序方便输出char tmp=charSet[i];charSet[i] = charSet[(valueCharacters-1)-i];charSet[(valueCharacters-1)-i] = tmp;}curOutPutTime= 0;maxOutPutTime = charArray[0].charCount;pausePoint = charArray[(valueCharacters-1)].charCount;while(curOutPutTime < maxOutPutTime){if(curOutPutTime == pausePoint){//w = valueCharacters;if(valueCharacters > 1){while(charArray[(valueCharacters-1)].charCount==pausePoint){int left=0,right = (valueCharacters-1);int index = ((left+right)/2);char expectValue = charArray[valueCharacters-1].charValue;while(left<=right){//二分查找if(charSet[index] == expectValue){for( k=index;k<(valueCharacters);k++)charSet[k]=charSet[k+1];break;}else if(charSet[index] < expectValue){left=index+1;index = (left+right)/2;}else if(charSet[index] > expectValue){right =index-1;index = (left+right)/2;}}//charSet[j]= 0 ;valueCharacters--;}if(valueCharacters > 0)pausePoint = charArray[valueCharacters-1].charCount;charSet[valueCharacters] = '\0';}}cout<<(char*)(charSet);curOutPutTime++;}cout<<endl;flush(cout);/*char chartest[]="again";cout<<chartest<<endl;*/again:continue;}};