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.
Tags :Two Pointers String
Solution 1 : my weak suolution, create a new string to input only alphanumeric
bool isPalindrome(string s) { if(s.empty())return true; string::iterator iter1=s.begin(),iter2=s.end(); string news; for(;iter1!=iter2;iter1++){ if(isalnum(*iter1)) news.push_back(tolower(*iter1)); } if(news.size()==1)return true; for(int i=0,j=news.size()-1;i<j;i++,j--){ if(news[i]!=news[j])return false; } return true; }
Solution 2: clean code
bool isPalindrome(string s) { for (int i = 0, j = s.size() - 1; i < j; i++, j--) { // Move 2 pointers from each end until they collide while (isalnum(s[i]) == false && i < j) i++; // Increment left pointer if not alphanumeric while (isalnum(s[j]) == false && i < j) j--; // Decrement right pointer if no alphanumeric if (toupper(s[i]) != toupper(s[j])) return false; // Exit and return error if not match } return true; }
时间: 2024-10-13 06:11:15