125. Verify palindrome (java implementation) --LeetCode

topic:

125. Verify palindrome

Given a string, verify whether it is a palindrome, only consider letters and numbers, and ignore the case of letters.

Explanation: In this question, we define an empty string as a valid palindrome string.

Example 1:

输入: "A man, a plan, a canal: Panama"
输出: true

Example 2:

输入: "race a car"
输出: false

Solution 1: Violence

/**
 * 思路：
 * 过滤出需要的字符
 * 反转字符串
 * 比较反转前后的字符串
 */
     public boolean isPalindrome(String s) {
    
    
        String filterS=isLetterORDigit(s);
        String reversS=reverse(filterS);
        return filterS.equalsIgnoreCase(reversS);
    }

    private String reverse(String filterS) {
    
    
        StringBuilder sb = new StringBuilder(filterS);
        sb.reverse();
        return sb.toString();
    }

    private String isLetterORDigit(String s) {
    
    
        return s.replaceAll("[^0-9a-zA-Z]","");
    }

Time complexity: On

Space complexity: On

Solution 2: Double pointer

/**
 * 思路：
 * 过滤出需要的字符
 * 头尾比较
 */
pub
    public boolean isPalindrome(String s) {
    
    
        StringBuilder sb = new StringBuilder();
        for (int i=0;i<s.length();i++){
    
    
            char c = s.charAt(i);
            if (Character.isLetterOrDigit(c)){
    
    
                sb.append(Character.toLowerCase(c));
            }
        }
        int l=0,r=sb.length()-1;
        while (r>l){
    
    
            if (sb.charAt(l++)!=sb.charAt(r--))return false;
        }
        return true;
    }

Time complexity: On

Space complexity: On