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125. Valid Palindrome - Easy

时间:2018-12-18 10:07:27      阅读:121      评论:0      收藏:0      [点我收藏+]

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

 

two pointer

注意比较char是否相同的时候先取出char再转换大小写,否则会超时

time: O(n), space: O(1)

class Solution {
    public boolean isPalindrome(String s) {
        int p1 = 0, p2 = s.length() - 1;
        while(p1 <= p2) {
            if(!Character.isLetterOrDigit(s.charAt(p1))) {
                p1++;
            }
            else if(!Character.isLetterOrDigit(s.charAt(p2))) {
                p2--;
            }
            else {
                if(Character.toLowerCase(s.charAt(p1)) != Character.toLowerCase(s.charAt(p2)))
                    return false;
                p1++;
                p2--;
            }
        }
        return true;
    }
}

 

125. Valid Palindrome - Easy

原文:https://www.cnblogs.com/fatttcat/p/10134964.html

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