Leetcode 125. Valid palindrome String

Valid palindrome

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:
Are you consider that the string might is empty? This was a good question to ask during a interview.

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

Main topic:

Test of Palindrome.

Ideas:

1. Clean the string and get the string with only numbers and letters.

2. Judge by comparing the characters to the end.

Class solution {public:    vector<string> stringsplit (String s,  const char * split)     {    vector<string>  result;    const int slen = s.length ();     char *cs = new char[slen + 1];    strcpy (Cs, s.data ());     char *p;        p = strtok (CS,  split);    while  (p)     {    printf ("%s \ n ",  p);     string tmp (P);     result.push_back (TMP);     p = strtok (Null, split);    }     Return result;    }    bool ispalindrome (String s)  {    if  (s.size ()  == 0 | |  s.size ()  == 1)     return true;    vector< String> vecstrs = stringsplit (S, " [email protected]#$%^&* ().,:;-?\" ");     s =  "";    for  (int i = 0; i  < vecstrs.size ();  i++)     s += vecStrs[i];     if  (s.size ()  == 1 | |  s.size ()  == 0)     return true;    int i  = 0;    for  (;  i < s.size ()  / 2; i++)     {    if  (s[i] <= 57 | |   s[s.size ()  - i - 1] <= 57)     {     if  (S[i] == s[s.size ()  - i -  1])     {    continue;    }     else    {    return false;    }     }    else if  (S[i] == s[s.size ()  - i  - 1] | |     s[i] - s[s.size ()  - i - 1] == 32 | |     s[s.size ()  - i - 1] - s[i] == 32)      {    continue;    }    else     {    return false;    }     }    return true;    }};

The above approach is inefficient and unfamiliar to the API.

Here are the improvements to the above:

Reference https://discuss.leetcode.com/topic/48376/12ms-c-clean-solution

The code is as follows:

Class Solution {Public:bool Ispalindrome (string s) {int i = 0, j = s.size ()-1;while (I < J) {while (!isalnum (S[i]) &am p;& i < J) I++;while (!isalnum (s[j]) && i < J) J--;if (ToLower (s[i++])! = ToLower (s[j--])) return false; return true;}};

Here, the isalnum () function is used to determine whether a literal number.

By using ToLower () to unify the case of a character, it becomes lowercase.

2016-08-11 13:26:25

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Leetcode 125. Valid palindrome String