Valid Palindrome

描述

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.

分析

无

代码

Python

// Valid Palindrome
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public boolean isPalindrome(String s) {
        s = s.toLowerCase();
        int left = 0;
        int right = s.length() - 1;
        while (left < right) {
            if (!Character.isLetterOrDigit(s.charAt(left))) ++left;
            else if (!Character.isLetterOrDigit(s.charAt(right))) --right;
            else if (s.charAt(left) != s.charAt(right)) return false;
            else { ++left; --right; }
        }
        return true;
    }
}

Java

// Valid Palindrome
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    bool isPalindrome(string s) {
        transform(s.begin(), s.end(), s.begin(), ::tolower);
        auto left = s.begin(), right = prev(s.end());
        while (left < right) {
            if (!::isalnum(*left))  ++left;
            else if (!::isalnum(*right)) --right;
            else if (*left != *right) return false;
            else { left++, right--; }
        }
        return true;
    }
};

C++

# Valid Palindrome
# 时间复杂度O(n)，空间复杂度O(1)
def isPalindrome(s: str) -> bool:
    s = s.lower()
    left = 0
    right = len(s) - 1
    while left < right:
        if not s[left].isalnum():
            left += 1
        elif not s[right].isalnum():
            right -= 1
        elif s[left] != s[right]:
            return False
        else:
            left += 1
            right -= 1
    return True

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