[LeetCode]--125. 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.

可以忽略除数字和字母以外的标点符号。

public boolean isPalindrome(String s) { if (s.length() == 0) return true; int i = 0, j = s.length() - 1; while (i < j) { while (i < j && !((s.charAt(i) >= 'a' && s.charAt(i) <= 'z') || (s.charAt(i) >= 'A' && s.charAt(i) <= 'Z') || (s .charAt(i) >= '0' && s.charAt(i) <= '9'))) i++; while (i < j && !((s.charAt(j) >= 'a' && s.charAt(j) <= 'z') || (s.charAt(j) >= 'A' && s.charAt(j) <= 'Z') || (s .charAt(j) >= '0' && s.charAt(j) <= '9'))) { System.out.println("--" + j); j--; } if (!(s.charAt(i) + "").equalsIgnoreCase(s.charAt(j) + "")) { return false; } i++; j--; } return true; }

这样写通过了。下面学习下别人的。

public class Solution { public boolean isPalindrome(String s) { if (s == null || s.length() == 0) { return true; } int front = 0; int end = s.length() - 1; while (front < end) { while (front < s.length() && !isvalid(s.charAt(front))){ // nead to check range of a/b front++; } if (front == s.length()) { // for emtpy string “.,,,” return true; } while (end >= 0 && ! isvalid(s.charAt(end))) { // same here, need to check border of a,b end--; } if (Character.toLowerCase(s.charAt(front)) != Character.toLowerCase(s.charAt(end))) { break; } else { front++; end--; } } return end <= front; } private boolean isvalid (char c) { return Character.isLetter(c) || Character.isDigit(c); } } 相关资源：敏捷开发V1.0.pptx