[回文数]
这道题很简单, 直接用两个字符数组保存回文数的左右部分加以判断就可以了
下面是部分代码.
bool palindrome_optimize(char* letter, int lenth) {char left[MaxLenth] = "";char right[MaxLenth] = "";int ml = 0, mr = 0;for (int i = 0, j = lenth - 1; i < lenth / 2& j >= lenth / 2; i++, j--) {left[ml++] = letter[i];right[mr++] = letter[j];}if (strncmp(left, right, lenth / 2) == 0) {return true;}else {return false;}
}