我能够反转一个字符串。例如，我可以反转＆＃34;反转字符串＆＃34;去＆＃34; esrever a gnirts＆＃34;。但是我无法一字一句地反转它，例如＆＃34;字符串反向＆＃34;。
void reverseString(char string[],char *start, char* end)
{
char tmp; //temporary variable to swap values
int count = 0;
while(start<end)
{
if(*start==' ')
{
printf("found space count %d \n",count);
reverseString(string,start-count,start);
}
tmp = *start;
*start = *end;
*end = tmp;
*start++;
*end--;
count++;
}
printf(" string %s \n", string);
}
int main()
{
char string[] = "reverse a string word by word";
char *start =string;
char *end =start+ strlen(string) -1;
reverseString(string,start,end);
return 0;
}
答案 0 :(得分：3)
做你已经完成的事情，然后反转整个结果（不专门处理空格）。
答案 1 :(得分：2)
这就是方法。我能够以字为单位反转字符串，以及整个字符串。只需查看代码，看看逻辑是否有帮助。
#include <stdio.h>
#include <string.h>
void stringrev(char *);
void reverseWords(char *);
void reverseString(char* , int);
int main()
{
char string[] = "reverse a string word by word";
reverseWords(string);
printf("\nWord-Wise Reversed String : %s\n",string);
stringrev(string);
return 0;
}
void reverseWords(char * str)
{
int i = 0, j = 0;
reverseString( str, strlen(str) );
while( 1 ) // Loop forever
{
if( *(str+j) == ' ' || *(str+j) == '\0') // Found a word or reached the end of sentence
{
reverseString( str+i, j-i );
i = j+1;
}
if( *(str+j) == '\0')
{
break;
}
j++;
}
}
void reverseString(char* str, int len)
{
int i, j;
char temp;
i=j=temp=0;
j=len-1;
for (i=0; i<j; i++, j--)
{
temp=str[i];
str[i]=str[j];
str[j]=temp;
}
}
void stringrev(char *str)
{
int i=-1,j=0;
char rev[50];
while(str[i++]!='\0');
while(i>=0)
rev[j++] = str[i--];
rev[j]='\0';
printf("\nComplete reverse of the string is : %s\n",rev);
}
答案 2 :(得分：1)
对此问题使用堆栈实现
步骤1：将字符串写入文件
步骤2：从文件中读取此内容并推送到链接列表
步骤3：在此链接列表上使用堆栈实现
Step4：从头到尾弹出链表!!
反过来了...... !!
答案 3 :(得分：0)
效率不高，但应该有效：
void reverse_string_word(char *data)
{
char *saveptr;
char *word;
char *tmp = malloc(strlen(data) + 1);
char *tmp2 = malloc(strlen(data) + 1);
*tmp = 0;
*tmp2 = 0;
word = strtok_r(data, " ", &saveptr);
if (word)
{
strcpy(tmp, word);
}
while (word)
{
word = strtok_r(NULL, " ", &saveptr);
if (word)
{
sprintf(tmp2, "%s %s", word, tmp);
strcpy(tmp, tmp2);
}
}
strcpy(data, tmp);
free(tmp);
free(tmp2);
}
答案 4 :(得分：0)
将其拆分为一个单词数组（char **），将其反转然后重新连接。
答案 5 :(得分：0)
这是一个不使用扫描程序或堆栈来解析单词的Java解决方案。从String的末尾开始并向后工作。可能更优雅但有效 - 如果有人有一个递归的java解决方案，我希望看到它。
"one two three four" is returned as "four three two one"
private void reverseWordsNoStackNoScanner(String str) {
System.out.println("reverseWordsNoStackNoScanner "+str);
String[] buff = new String[str.length()];
int end=str.length()-1;
int j=end;
int start=0;
int ptr=0;
for (int i=str.length()-1;i>=0;i--){
boolean writeBuff=false;
if (str.charAt(i)!=' ') { // have we backed up to a blank?
j--; //no
} else { //yes! write out this word
writeBuff=true;
}
if (i==0) writeBuff=true; //are we done (position 0)?
if (writeBuff) { //time to write a word?
//we've hit a delimiter (or we're done)
ptr=j; //pointing at a blank or the beginning
ptr++; //bump past the blank or the beginning
while(ptr<=end){ //write the word from beginning to finish
buff[start++]=String.valueOf(str.charAt(ptr++));
}
//don't write a blank when we are on the last word (past the end)
if (i>0)buff[start++]=" ";
j--;
//set pointers for next iteration
end=j; //back up end ptr to new 'end' - to parse the next word
}
}
//print out our reversed word string
for (String s: buff) {
System.out.print(s);
}
}
答案 6 :(得分：0)
从开头读取字符串并将每个字符推入字符堆栈。一旦遇到空格或换行符或eof，开始弹出字符并在标准输出上逐个打印或根据需要将其存储在文件中。
答案 7 :(得分：0)
'My name is' return as 'yM eman si'
'My name is' return as 'yM eman si'
#include<stdio.h>
#include<malloc.h>
#include<string.h>
void reve(char [],int,int);
void main()
{
char *a;
int i,j=-1;
a=(char*)malloc(1000*sizeof(char));
gets(a);
for(i=0;a[i]!='\0';i++)
{
if(a[i]!=' '&&j==-1)
j=i;
if((a[i]==' '&&a[i+1]!=' '&&j>=0)||a[i+1]=='\0')
{
a[i]==' '?reve(a,j,i-1):reve(a,j,i);
j=-1;
}
}
for(i=0;a[i]!='\0';i++)
printf("%c",a[i]);
}
void reve(char a[],int j,int i)
{
char temp;
if(a[0]==' ')
j=0;
for(;i!=j&&j<i;i--,j++)
{
temp=a[j];
a[j]=a[i];
a[i]=temp;
}
}
答案 8 :(得分：0)
<table>
<tr>
<td>1</td>
<td>2</td>
</tr>
<tr>
<td>3</td>
<td>4</td>
</tr>
</table>
<p>In IE 10+ and Edge: the red background of the table leaks through the bottom of the cells when border-radius is applied</p>}
'i love india' return as 'india love i'
' i love india' return as 'india love i'
#include<stdio.h>
#include<malloc.h>
#include<string.h>
void reve(char [],int,int);
void main()
{
char *a;
int i,j=-1;
a=(char*)malloc(1000*sizeof(char));
gets(a);
for(i=0;a[i]!='\0';i++)
{
if(a[i]!=' '&&j==-1)
j=i;
if((a[i]==' '&&a[i+1]!=' '&&j>=0)||a[i+1]=='\0')
{
a[i]==' '?reve(a,j,i-1):reve(a,j,i);
j=-1;
}
}
reve(a,0,i-1);
for(i=0;a[i]!='\0';i++)
printf("%c",a[i]);
}
答案 9 :(得分：0)
使用C＃反转String中的单词
public string ReverseWordsInString(string inputString)
{
string output = string.Empty;
string[] splitStrings = inputString.Split(' ');
for (int i = splitStrings.Length-1; i > -1 ; i--)
{
output = output + splitStrings[i]+ " ";
}
return output;
}
答案 10 :(得分：0)
它不应该像其他人所发布的那样复杂，请参考并提供反馈，很高兴对其进行改进。
let
Source = SharePoint.Tables("https://sharepoint2013.contoso.com/sites/201O9A/", [ApiVersion = 15]),
#"ddasddc8c-54b1-4bc2-a05d-3dasde758392" = Source{[Id="ddasddc8c-54b1-4bc2-a05d-3dasde758392"]}[Items]
in
#"ddasddc8c-54b1-4bc2-a05d-3dasde758392"
输出：“反向字符串”