每天一道C\C++笔试题IV---翻转字符串
这也是笔试中一道经典的C语言题：
给定一个字符串，将其翻转。如abc ==> cba
拿到此题时，我是想都没想，直接说，再用一个字符串tmp来缓存一下此串，然后一个for循环赋值搞定。
思路有了，代码就有了。
[cpp]
#include <stdio.h>
#include <stdlib.h>
int main()
{
char string[20],tmp[20];
int length;
printf("please input less than 20 char:");
scanf("%s",string);
printf("your input string is %s\n",string);
length = strlen(string);
printf("length is %d\n",length);
for(int i = 0;i<length;++i)
{
tmp[i] = string[i];
}
for(int i = 0;i<length;++i)
{
string[i] = tmp[length-i-1];
}
printf("after revert:%s\n",string);
return 0;
}
用gcc编译：gcc -o revert revert_string.c -std=c99
后运行，结果如我所料。但是我这个算法太不优雅了，因为将两个字符数组赋值就用了一个循环，然后翻转时再用一个循环，这效率真是不敢恭维。
优雅的方法应该是这样的思路：
找到这个字符串的中间位置，然后将其左边的字符与右边的字符交换位置。
实现起来应该是下面这样：
[cpp]
#include <stdio.h>
#include <stdlib.h>
int main()
{
char string[20], tmp;
int length;
printf("please input less than 20 char:");
scanf("%s",string);
printf("your input string is %s\n",string);
//get string length,very useful method
for(length=0;string[length];length++)
;
printf("length is %d\n",length);
//very beateful !!!
for(int i=0;i<length/2;i++)
{
tmp = string[i];
printf("tmp is %c\n" ,string[i]);
string[i] = string[length-i-1];
printf("string[%d] is %c\n",i,string[length-i-1]);
string[length-i-1] = tmp;
printf("string[%d] is %c\n",length-i-1,tmp);
}
printf("after revert:%s\n",string);
return 0;
}
运行效果如下：
[plain]
D:\workspace\C\revert_string>gcc -o revert revert_string.c -std=c99
D:\workspace\C\revert_string>revert
please input less than 20 char:abc
your input string is abc
length is 3
tmp is a
string[0] is c
string[2] is a
after revert:cba
D:\workspace\C\revert_string>revert
please input less than 20 char:abcd
your input string is abcd
length is 4
tmp is a
string[0] is d
string[3] is a
tmp is b
string[1] is c
string[2] is b
after revert:dcba
这样的算法，相比之前要提高甚多效率。只要开动脑筋，世界会更加优雅。
补充：软件开发 , C++ ,
