我一直在从事一项作业，要求输入包含由小写字母字符组成的单词，并用空格（或换行）分隔。它以单词“ STOP”终止。您可以假设不超过100个单词，每个单词的长度不超过20个字符。
输出5个最大的字谜组。如果少于5组，则全部输出。通过减小大小对组进行排序。在字典上按字典最小元素断开关系。对于每个组输出，请打印其大小和成员词。按字典顺序对成员词进行排序，并且仅打印一次相等词。
#include <stdio.h>
#include <string.h>
void ReadWord(char [21]);
void SortWord(char [21]);
void AssignAnagramGroup(int);
void SortWordList();
void CalculateTopFiveGroups();
int isAnagram(char [21], char [21]);
char wordlist[100][21]; //stores list of words
int anagramgroup[100]; //for each word, store the group id it belongs to
int anagramgroupcount=1; //how many groups do we have
int anagramgroupsize[101]; //for each anagramgroup, we store size
int wordcount=0; //how many words were in the list
int largestfive[5]={0}; //stores top 5 anagramgroups
int groupsizeorder[100]; //array to hold the groupsize in corresponding word placement
int main ()
{
char word[21];
int done = 0;
int i, j, k;
int x=1;
do
{
ReadWord(word);
if (strcmp(word, "STOP")==0 || wordcount > 99)
break;
strcpy(wordlist[wordcount++], word);
AssignAnagramGroup(wordcount-1);
}while(!done);
for (i=0; i<wordcount; i++){
printf ("ReadWord: Wordlist[%d]=\"%s\" AnagramGroup=%d\n", i, wordlist[i], anagramgroup[i]);}
printf ("WordList size=%d words\n", wordcount);
for (i=1; i<=5; i++)
if (anagramgroupsize[i]!=0)
printf ("Anagram Group %d of size %d\n", i, anagramgroupsize[i]);
printf ("\n");
SortWordList();
CalculateTopFiveGroups();
for (i=0; i<wordcount; i++){
printf ("Wordlist[%d]=\"%s\" Gid=%d Gsize=%d\n", i, wordlist[i], anagramgroup[i], groupsizeorder[i]);}
//THIS IS WHAT NEEDS CHANGING****
for (i=0; i<5 && largestfive[i+1]!=0; i++){
printf("Group size of: %d ", largestfive[i+1]);
printf("\n");
}
return 0;
}
void ReadWord(char word [])
{
scanf ("%s", word);
}
void AssignAnagramGroup (int wordcount)
{
int j;
if (wordcount==0)
{
anagramgroup[0]=1;
anagramgroupsize[1]=1;
}
else
{
for (j=0; j<wordcount; j++)
{
if (isAnagram(wordlist[wordcount], wordlist[j]) ==1)
{
anagramgroup[wordcount]=anagramgroup[j];
anagramgroupsize[anagramgroup[j]]+=1;
break;
}
}
if (j==wordcount){
anagramgroup[wordcount]=++anagramgroupcount;
anagramgroupsize[anagramgroup[j]]+=1;
}
}
return;
}
int isAnagram (char word1[],char word2[])
{
int first[26] = {0}, second[26] = {0}, c = 0;
while (word1[c] != '\0')
{
first[word1[c]-'a']++;
c++;
}
c = 0;
while (word1[c] != '\0')
{
second[word2[c]-'a']++;
c++;
}
for (c = 0; c < 26; c++)
{
if (first[c] != second[c])
return 0;
}
return 1;
}
void SortWordList()
{
int i, j, k, temp;
char tempchar[21];
for (i=0; i<wordcount; i++) {
for (j=0; j<wordcount-1; j++) {
if (strcmp(wordlist[j], wordlist[j+1])>0){
strcpy(tempchar, wordlist[j]);
strcpy(wordlist[j], wordlist[j+1]);
strcpy(wordlist[j+1], tempchar);
temp=anagramgroup[j];
anagramgroup[j]=anagramgroup[j+1];
anagramgroup[j+1]=temp;
}
}
}
for (i=0; i<wordcount; i++)
groupsizeorder[i]=anagramgroupsize[anagramgroup[i]];
return;
}
void CalculateTopFiveGroups()
{
int i, j, temp;
int anagramtemp[101]={0};
for (i=1; i<=anagramgroupcount; i++){
anagramtemp[i]=anagramgroupsize[i];
}
for (i=0; i<anagramgroupcount; i++) {
for (j=0; j<anagramgroupcount-i; j++) {
if (anagramtemp[j+1]<anagramtemp[j+2]){
temp=anagramtemp[j+1];
anagramtemp[j+1]=anagramtemp[j+2];
anagramtemp[j+2]=temp;
}
}
}
for (i=1; i<=5; i++){
largestfive[i]=anagramtemp[i];
}
return;
}
我目前有输出
ReadWord: WordList[0]="ate" AnagramGroup=1
ReadWord: WordList[1]="eta" AnagramGroup=1
ReadWord: WordList[2]="colli" AnagramGroup=2
ReadWord: WordList[3]="icoll" AnagramGroup=2
ReadWord: WordList[4]="eat" AnagramGroup=1
ReadWord: WordList[5]="lloci" AnagramGroup=2
ReadWord: WordList[6]="cat" AnagramGroup=3
ReadWord: WordList[7]="tac" AnagramGroup=3
WordList size=8 words
Anagram Group 1 of size 3
Anagram Group 2 of size 3
Anagram Group 3 of size 2
WordList[0]="ate" Gid=1 Gsize=3
WordList[1]="cat" Gid=3 Gsize=2
WordList[2]="colli" Gid=2 Gsize=3
WordList[3]="eat" Gid=1 Gsize=3
WordList[4]="eta" Gid=1 Gsize=3
WordList[5]="icoll" Gid=2 Gsize=3
WordList[6]="lloci" Gid=2 Gsize=3
WordList[7]="tac" Gid=3 Gsize=2
Group of size 3
Group of size 3
Group of size 2
这就是我需要包含的输出
ReadWord: WordList[0]="ate" AnagramGroup=1
ReadWord: WordList[1]="eta" AnagramGroup=1
ReadWord: WordList[2]="colli" AnagramGroup=2
ReadWord: WordList[3]="icoll" AnagramGroup=2
ReadWord: WordList[4]="eat" AnagramGroup=1
ReadWord: WordList[5]="lloci" AnagramGroup=2
ReadWord: WordList[6]="cat" AnagramGroup=3
ReadWord: WordList[7]="tac" AnagramGroup=3
WordList size=8 words
Anagram Group 1 of size 3
Anagram Group 2 of size 3
Anagram Group 3 of size 2
WordList[0]="ate" Gid=1 Gsize=3
WordList[1]="cat" Gid=3 Gsize=2
WordList[2]="colli" Gid=2 Gsize=3
WordList[3]="eat" Gid=1 Gsize=3
WordList[4]="eta" Gid=1 Gsize=3
WordList[5]="icoll" Gid=2 Gsize=3
WordList[6]="lloci" Gid=2 Gsize=3
WordList[7]="tac" Gid=3 Gsize=2
Group of size 3:[GID: 1] ate eat eta
Group of size 3:[GID: 2] colli icoll lloci
Group of size 2:[GID: 3] cat tac
我很确定我需要更改main上最后一个标有// //这是需要更改的内容的打印****
如果有人不能帮助我，并弄清楚如何使最后一个打印声明正确无误，我将不胜感激。
最佳答案
这适用于您的特定示例，但是在使用其他示例时会出现一些错误。也许会有所帮助。
`do {
ReadWord(word);
if (strcmp(word, "STOP") == 0 || wordcount > 99) // if STOP is entered or 100 words have been read
break; // the input loop will be exited
strcpy(wordlist[wordcount++], word); // assigns words read in to the word list
AssignAnagramGroup(wordcount-1); // assigns word read in to an anagram group
} while (!done); // forever loop, only exits on break condition
SortWordList(); // sorts the word list
TopFiveGroups(); // finds the top five largest groups
for (i=0; i<wordcount; i++) {
printf ("Wordlist[%d] = \"%s\" AnagramGroup= %d, Gsize=%d\n", i, wordlist[i], anagramgroup[i], groupsizeorder[i]);
}
printf("Wordlist size = %d words \n", wordcount);
for (i=0; i<5 && largest[i+1] != 0; i++) { // prints other stuff
printf("Anagram Group %d of size %d ", i+1, largest[i+1]);
for (j=0; j<wordcount; j++) // prints words
if (anagramgroup[j] == i+1)
printf("%s ", wordlist[j]);
printf("\n");
}
return 0;`
关于c - 给定一些文本输入，找到最大的字谜组，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/35808492/