LeetCode - Valid Palindrome

Valid Palindrome

2014.1.13 18:48

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

Solution:

　　This problem doesn't involve maths or difficult algorithms, so it's the kind of easy problem to test if you're careful enough to solve it one-pass. Make NO mistake.

　　Time complexity is O(n), where n is the length of the string. Space complexity is O(1).

Accepted code:

 1 // 1RE, 1AC, be more careful, could've 1AC~

 2 class Solution {

 3 public:

 4     bool isPalindrome(string s) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         int i, j, len;

 8         

 9         len = s.length();

10         if(len <= 0){

11             return true;

12         }

13         

14         char a, b;

15         

16         i = 0;

17         j = len - 1;

18         while(i < j){

19             if(s[i] >= 'a' && s[i] <= 'z'){

20                 a = s[i];

21             }else if(s[i] >= '0' && s[i] <= '9'){

22                 a = s[i];

23             }else if(s[i] >= 'A' && s[i] <= 'Z'){

24                 a = s[i] - 'A' + 'a';

25             }else{

26                 ++i;

27                 continue;

28             }

29             if(s[j] >= 'a' && s[j] <= 'z'){

30                 b = s[j];

31             }else if(s[j] >= '0' && s[j] <= '9'){

32                 b = s[j];

33             }else if(s[j] >= 'A' && s[j] <= 'Z'){

34                 b = s[j] - 'A' + 'a';

35             }else{

36                 // 1RE here, wrong direction of $j

37                 --j;

38                 continue;

39             }

40             if(a == b){

41                 ++i;

42                 --j;

43             }else{

44                 break;

45             }

46         }

47         

48         return i >= j;

49     }

50 };