問題描述
編寫一個函數,其作用是將輸入的字符串反轉過來。輸入字符串以字符數組 char[]
的形式給出。
不要給另外的數組分配額外的空間,你必須原地修改輸入數組、使用 O(1) 的額外空間解決這一問題。
你可以假設數組中的所有字符都是 ASCII 碼錶中的可打印字符。
示例 1:
輸入:["h","e","l","l","o"]
輸出:["o","l","l","e","h"]
示例 2:
輸入:["H","a","n","n","a","h"]
輸出:["h","a","n","n","a","H"]
代碼一 :利用臨時變量,頭尾交換,一直到兩者位置相等
void reverse(char *str, int len)
{
int i = 0;
int j = len - 1;
int tmp;
while (i < j)
{
tmp = str[i];
str[i++] = str[j];
str[j--] = tmp;
}
}
int main()
{
char str[] = { 'h','l','o','o' };
int len = sizeof(str) / sizeof(str[0]);
reverse(str, len);
for each (char val in str)
{
cout << val << " ";
}
cout << endl;
for (char val : str)
{
cout << val << " ";
}
cout << endl;
getchar();
return 0;
}
代碼二:利用STL容器並調用其相關函數
int main()
{
vector<char>str = { 'h','l','o','o' };
reverse(str.begin(), str.end());
for each (char val in str)
{
cout << val << " ";
}
cout << endl;
for (char val : str)
{
cout << val << " ";
}
getchar();
return 0;
}