【LeetCode】5. Valid palindrome

Title Description  

Description in English

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.

English address

https://leetcode.com/problems/valid-palindrome/https://leetcode.com/problems/valid-palindrome/

Description of Chinese version

Given a string s , verification s Whether it is Palindrome string , Consider only alphabetic and numeric characters , The case of letters can be ignored . In this question , Define an empty string as a valid Palindrome string .

Example 1: Input : s = "A man, a plan, a canal: Panama" Output : true explain ："amanaplanacanalpanama" It's a palindrome string Example 2: Input : s = "race a car" Output : false explain ："raceacar" It's not a palindrome string

Constraints· Tips ：

• 1 <= s.length <= 2 * 105

• character string s from ASCII Character composition

Address of Chinese version

https://leetcode.cn/problems/XltzEq/https://leetcode.cn/problems/XltzEq/

Their thinking

First turn all lowercase letters （ The title requires that case be ignored ） Then remove the characters except letters and numbers , Compare first and last

How to solve the problem

My version

class Solution {
 public static boolean isPalindrome(String s) {
        String s1 = s.toLowerCase().replaceAll("[^a-z|0-9]", "");
        int length = s1.length();
        for (int i = 0; i < length; i++) {
            int j = length - i - 1;
            if(j <= i){
                break;
            }
            char c1 = s1.charAt(i);
            char c2 = s1.charAt(j);
            if (c1 != c2) {
                return false;
            }
        }
        return true;
    }
}

Official edition

Screening + Compare

class Solution {
        public boolean isPalindrome(String s) {
        StringBuffer sgood = new StringBuffer();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char ch = s.charAt(i);
            if (Character.isLetterOrDigit(ch)) {
                sgood.append(Character.toLowerCase(ch));
            }
        }
        StringBuffer sgood_rev = new StringBuffer(sgood).reverse();
        return sgood.toString().equals(sgood_rev.toString());
    }
}

Double finger needling

class Solution {
    public boolean isPalindrome(String s) {
        StringBuffer sgood = new StringBuffer();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char ch = s.charAt(i);
            if (Character.isLetterOrDigit(ch)) {
                sgood.append(Character.toLowerCase(ch));
            }
        }
        int n = sgood.length();
        int left = 0, right = n - 1;
        while (left < right) {
            if (Character.toLowerCase(sgood.charAt(left)) != Character.toLowerCase(sgood.charAt(right))) {
                return false;
            }
            ++left;
            --right;
        }
        return true;
    }
}

summary

My method seems to be very similar to the official double pointer method , But the execution time and memory consumption are more , Especially the execution time 》〉 As expected, there is no harm without comparison (╯﹏╰）

At first, I thought it was the one I used in the later comparison String, Official StringBuffer,StringBuffer Than String The reason why traversal is fast , So I assigned the regular matching string in my code to a StringBuffer, And then use that StringBuffer Traversal , But it is of no damn use (-｡-; More slowly ...

​But！ When I replace regular matching with Character.isLetterOrDigit(), The speed is soaring ～～

The regular matching of the original string is so slow , Welcome students who know the reason to kick me in the comment area *〜（ゝ.∂）