写一个函数，实现翻转字符串的功能，并考虑时间复杂度、空间复杂度，使其尽可能最小

1. 字符串翻转：请把"I am a student."，翻转成".tneduts a ma I"；

吾来打个样：

#include <stdio.h>
void reverseCharArray(char *str, char *begin, char *end) {while(begin < end) { //取地址，交换两个字符char temp = *begin;*begin = *end;*end = temp;begin++;end--;}
}char *reverse_str(char *str) {if(NULL == str) { //字符串为空直接返回；return str;}char *begin;char *end;begin = end = str;while(*end != '\0') { //end指向字符串的末尾；end++;}--end;  //取出最后一个'.'字符；//取字符数组下标，进行交换；reverseCharArray(str, begin, end);return str; //返回结果；
}int main(int argc, const char * argv[]) {char str[] = "I am a student.";printf("%s\n", reverse_str(str));return 0;
}

2. 字符串翻转：请把 "I am a student."，翻转成 "I ma a .tneduts"；

吾来打个样：

#include <stdio.h>
void reverseCharArray(char *str, char *begin, char *end) {while(begin < end) { //取地址，交换两个字符char temp = *begin;*begin = *end;*end = temp;begin++;end--;}
}char *reverseChar(char *str) {char *begin;char *end;begin = end = str;while (*end != '\0') {if (*end == ' ') {reverseCharArray(str, begin, end - 1);begin = end + 1;}else if (*(end + 1) == '\0') {  //当*(end + 1)取出的是字符串数组最后的'\0'的时候reverseCharArray(str, begin, end);}end++;}return str;
}int main(int argc, const char * argv[]) {char str[] = "I am a student.";printf("%s\n", reverseChar(str));return 0;
}

3. 字符串翻转：请把 "I am a student."，翻转为 "student. a am i"；

吾来打个样：

#include <stdio.h>
void reverseCharArray(char *str, char *begin, char *end) {while(begin < end) { //取地址，交换两个字符char temp = *begin;*begin = *end;*end = temp;begin++;end--;}
}char *reverseChar(char *str) {char *begin;char *end;begin = end = str;while (*end != '\0') {if (*end == ' ') {reverseCharArray(str, begin, end - 1);begin = end + 1;}else if (*(end + 1) == '\0') {  //当*(end + 1)取出的是字符串数组最后的'\0'的时候reverseCharArray(str, begin, end);}end++;}return str;
}int main(int argc, const char * argv[]) {char str[] = "I am a student.";char *begin;char *end;begin = end = str;while(*end != '\0') { //end指向字符串的末尾；end++;}--end;  //取出最后一个'.'字符；reverseCharArray(str, begin, end);printf("%s\n", reverseChar(str));return 0;
}