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。

#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：

#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：

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。