# Valid Palindrome - tags: [palindrome] ### Source - leetcode: [Valid Palindrome | LeetCode OJ](https://leetcode.com/problems/valid-palindrome/) - lintcode: [(415) Valid Palindrome](http://www.lintcode.com/en/problem/valid-palindrome/) ~~~ Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. 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. Challenge O(n) time without extra memory. ~~~ ### 题解 字符串的回文判断问题，由于字符串可随机访问，故逐个比较首尾字符是否相等最为便利，即常见的『两根指针』技法。此题忽略大小写，并只考虑字母和数字字符。链表的回文判断总结见 [Check if a singly linked list is palindrome](http://algorithm.yuanbin.me/zh-cn/linked_list/check_if_a_singly_linked_list_is_palindrome.html). ### Python ~~~ class Solution: # @param {string} s A string # @return {boolean} Whether the string is a valid palindrome def isPalindrome(self, s): if not s: return True l, r = 0, len(s) - 1 while l < r: # find left alphanumeric character if not s[l].isalnum(): l += 1 continue # find right alphanumeric character if not s[r].isalnum(): r -= 1 continue # case insensitive compare if s[l].lower() == s[r].lower(): l += 1 r -= 1 else: return False # return True ~~~ ### C++ ~~~ class Solution { public: /** * @param s A string * @return Whether the string is a valid palindrome */ bool isPalindrome(string& s) { if (s.empty()) return true; int l = 0, r = s.size() - 1; while (l < r) { // find left alphanumeric character if (!isalnum(s[l])) { ++l; continue; } // find right alphanumeric character if (!isalnum(s[r])) { --r; continue; } // case insensitive compare if (tolower(s[l]) == tolower(s[r])) { ++l; --r; } else { return false; } } return true; } }; ~~~ ### Java ~~~ public class Solution { /** * @param s A string * @return Whether the string is a valid palindrome */ public boolean isPalindrome(String s) { if (s == null || s.isEmpty()) return true; int l = 0, r = s.length() - 1; while (l < r) { // find left alphanumeric character if (!Character.isLetterOrDigit(s.charAt(l))) { l++; continue; } // find right alphanumeric character if (!Character.isLetterOrDigit(s.charAt(r))) { r--; continue; } // case insensitive compare if (Character.toLowerCase(s.charAt(l)) == Character.toLowerCase(s.charAt(r))) { l++; r--; } else { return false; } } return true; } } ~~~ ### 源码分析 两步走： 1. 找到最左边和最右边的第一个合法字符(字母或者字符) 1. 一致转换为小写进行比较 字符的判断尽量使用语言提供的 API ### 复杂度分析 两根指针遍历一次，时间复杂度 O(n)O(n)O(n), 空间复杂度 O(1)O(1)O(1).