[LeetCode] 125. Valid Palindrome

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Constraints:

• 1 <= s.length <= 2 * 105
• s consists only of printable ASCII characters.

验证回文串。

如果在将所有大写字符转换为小写字符、并移除所有非字母数字字符之后，短语正着读和反着读都一样。则可以认为该短语是一个 回文串 。

字母和数字都属于字母数字字符。

给你一个字符串 s，如果它是 回文串 ，返回 true ；否则，返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/valid-palindrome
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题意很简单，验证给的 input 是否是一个有效的回文。思路也很简单，two pointer 逼近。注意

• 只比较 input 里面的字母，如果是数字或者空格就跳过
• 不需要区分大小写 - 把字母都变成小写再比较

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {string} s
 * @return {boolean}
 */
var isPalindrome = function (s) {
    if (s === null || s.length === 0) return true;
    const alphanum = s.toLowerCase().replace(/[\W]/g, "");
    let left = 0;
    let right = alphanum.length - 1;
    while (left < right) {
        if (alphanum[left] !== alphanum[right]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
};

 

Java实现

class Solution {
    public boolean isPalindrome(String s) {
        // corner case
        if (s == null || s.length() == 0) {
            return true;
        }

        // normal case
        int left = 0;
        int right = s.length() - 1;
        while (left < right) {
            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
                left++;
            }
            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
                right--;
            }
            if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

 

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