回文串之判断

 1 //从左右两边开始比较判断
 2 #include<iostream>
 3 #include<string.h>
 4 using namespace std;
 5 const int MAXN = 1e5;
 6 bool is_palindrome(char *s, int len)
 7 {
 8     if(s == NULL || len < 1)    //非法输入
 9         return false;
10     char *front, *tail;    //首尾指针
11     //指针声明了就要记得初始化
12     front = s;
13     tail = s + len - 1;
14     while(front < tail)
15     {
16         if(*front != *tail)
17             return false;
18         front++;
19         tail--;
20     }
21     return true;
22 }
23 int main()
24 {
25     char str[MAXN];
26     while(cin >> str)
27     {
28         int len = strlen(str);
29         cout << is_palindrome(str, len) << endl;
30     }
31     return 0;
32 }
33 //时间复杂度O(n),空间复杂度O(1)

 1 //从中间开始往两边判断
 2 #include<iostream>
 3 #include<string.h>
 4 using namespace std;
 5 const int MAXN = 1e5;
 6 bool is_palindrome(char *s, int len)
 7 {
 8     if(s == NULL || len < 1)
 9         return false;
10     char *left, *right;
11     int mid = ((len >> 1) - 1) >= 0 ? (len >> 1) - 1 : 0;
12     left = s + mid;
13     right = s + len - 1 - mid;
14     while(left >= s)
15     {
16         if(*left != *right)
17             return false;
18         left--;
19         right++;
20     }
21     return true;
22 }
23 int main()
24 {
25     char str[MAXN];
26     while(cin >> str)
27     {
28         int len = strlen(str);
29         cout << is_palindrome(str, len) << endl;
30     }
31     return 0;
32 }
33 //时间复杂度O(n),空间复杂度O(1)