问题描写:
写1个函数将i am a student 转化为 student a am i
解题思路:

//i am a student --->--> student a am i
#include<iostream>
using namespace std;
/*翻转函数--->利用交换*/
void words_reverse(char str[],int left,int right)
{
while(left < right)
{
str[left] += str[right];
str[right] = str[left] - str[right];
str[left] -= str[right];
left++;
right--;
}
}
int main()
{
char str[] = " student a am i";
/*找到数组的(最左侧)和(最右侧)后整体翻转*/
int left = 0;
int right = 0;
while(str[right] != '')
{
right++;
}
words_reverse(str,left,right⑴);
/*找到各个单词的(最左侧)和(最右侧),将各个单词分别翻转*/
int lowflag = 0; //状态标志位:要反转的字符串的左侧是不是找到
int highflag = 0;//状态标志位:要反转的字符串的右侧是不是找到
int i = 0;
while(str[i] != '')
{
/*寻觅单词的最左侧即低位*/
if(str[i] != ' ' && lowflag == 0)//在没有找到单词最左侧的条件下,如果发现1个非空格字符表示最左侧找到
{
lowflag = 1;
left = i;
}
/*if(str[i] == ' ' && highflag == 0)//char str[] = "i am a student "情况不适合(最后面有空格)
{
highflag = 1;
right = i⑴;
}*/
/*寻觅单词的最右侧即低位*/
if(lowflag == 1 && str[i] == ' ' && highflag == 0)//char str[] = "student a am i";情况不适合(第1个单词没翻转)
{
highflag = 1;//在没有找到单词最右侧的条件下,如果发现1个空格字符,则空格前1个是单词最右侧字符
right = i⑴;
}
/*最左侧(低位)和最右侧(高位)都找到了以后完成翻转*/
if(lowflag == 1 && highflag == 1)
{
words_reverse(str,left,right);
lowflag = 0;
highflag = 0;
}
i++;
}
cout<<str<<endl;
return 0;
}