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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 }
LeetCode-Valid Palindrome
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