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  • LeetCode-Valid Palindrome

    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.

    Solution:

     1 public class Solution {
     2     public boolean isPalindrome(String s) {
     3         if (s.isEmpty()) return true;
     4 
     5         int p1 = 0, p2 = s.length()-1;
     6      while (p1<s.length() && !( (s.charAt(p1)>='A' && s.charAt(p1)<='Z')  || (s.charAt(p1)>='a' && s.charAt(p1)<='z') || (s.charAt(p1)>='0' && s.charAt(p1)<='9') ) ) p1++;
     7     while (p2>=0 && !( (s.charAt(p2)>='A' && s.charAt(p2)<='Z')  || (s.charAt(p2)>='a' && s.charAt(p2)<='z') || (s.charAt(p2)>='0' && s.charAt(p2)<='9') ) ) p2--;
     8 
     9         while (p1<p2){
    10             char c1 = s.charAt(p1);
    11             char c2 = s.charAt(p2);
    12             if (c1 == c2 || Math.abs((int) c1 - (int) c2) == Math.abs( (int) 'A' - (int) 'a' ) ){
    13                 p1++;
    14                 p2--;
    15             while (p1<s.length() && !( (s.charAt(p1)>='A' && s.charAt(p1)<='Z')  || (s.charAt(p1)>='a' && s.charAt(p1)<='z') || (s.charAt(p1)>='0' && s.charAt(p1)<='9') ) ) p1++;
    16          while (p2>=0 && !( (s.charAt(p2)>='A' && s.charAt(p2)<='Z')  || (s.charAt(p2)>='a' && s.charAt(p2)<='z') || (s.charAt(p2)>='0' && s.charAt(p2)<='9') ) ) p2--;
    17        } else return false;
    18         }
    19 
    20         return true;
    21     }
    22 }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/4202654.html
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