每天一道C\C++筆試題IV—翻轉字串

這也是筆試中一道經典的C語言題：

給定一個字串，將其翻轉。如abc ==> cba

拿到此題時，我是想都沒想，直接說，再用一個字串tmp來緩衝一下此串，然後一個for迴圈賦值搞定。

思路有了，代碼就有了。

#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編譯：gcc -o revert revert_string.c -std=c99

後運行，結果如我所料。但是我這個演算法太不優雅了，因為將兩個字元數組賦值就用了一個迴圈，然後翻轉時再用一個迴圈，這效率真是不敢恭維。

優雅的方法應該是這樣的思路：

找到這個字串的中間位置，然後將其左邊的字元與右邊的字元交換位置。

實現起來應該是下面這樣：

#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 methodfor(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;}

運行效果如下：

D:\workspace\C\revert_string>gcc -o revert revert_string.c -std=c99D:\workspace\C\revert_string>revertplease input less than 20 char:abcyour input string is abclength is 3tmp is astring[0] is cstring[2] is aafter revert:cbaD:\workspace\C\revert_string>revertplease input less than 20 char:abcdyour input string is abcdlength is 4tmp is astring[0] is dstring[3] is atmp is bstring[1] is cstring[2] is bafter revert:dcba

這樣的演算法，相比之前要提高甚多效率。只要開動腦筋，世界會更加優雅。