[LeetCode] Valid Palindrome, Solution

最新推荐文章于 2024-01-04 12:53:14 发布

原创最新推荐文章于 2024-01-04 12:53:14 发布 · 185 阅读

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本文介绍了一种使用双指针从两端向中间遍历的方法来判断一个字符串是否为回文。该方法只考虑字母数字字符，并忽略大小写差异。此外，还讨论了如何处理空字符串的情况。

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.

» Solve this problem

[Thoughts]
Two pointer. From both sides to middle.

[Code]

1:       bool isPalindrome(string s) {  
2:            int start = 0;  
3:            int end = s.size()-1;  
4:            std::transform(s.begin(), s.end(), s.begin(), ::tolower);  
5:            while(start<end)  
6:            {  
7:                 while(start< end && !isAlpha(s[start])) start++;  //filter non-alpha char
8:                 while(start< end && !isAlpha(s[end])) end--;  //filter non-alpha char
9:                 if(s[start]!=s[end]) break;        
10:                 start++;  
11:                 end--;  
12:            }  
13:            if(start >= end)  
14:                 return true;  
15:            else  
16:                 return false;        
17:       }  
18:       bool isAlpha(char c)  
19:       {  
20:            if(c>='a' && c<='z') return true;     
21:            if(c>='0' && c<='9') return true;  
22:            return false;  
23:       }