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
判断字母数字字符的回文串,直接头尾反向比较,跳过其他字符
//java
class Solution {
public boolean isPalindrome(String s) {
if(s == null || s.length() == 0 ) return true;
int i = 0,j = s.length()-1;
for(i=0,j=s.length()-1;i<j;++i,--j){
while(i<j && !Character.isLetterOrDigit(s.charAt(i))) i++;
while(i<j && !Character.isLetterOrDigit(s.charAt(j))) j--;
if(i<j && Character.toLowerCase(s.charAt(i))!=Character.toLowerCase(s.charAt(j))) return false;
}
return true;
}
}
本文介绍了一种算法,用于判断一个字符串是否为回文,只考虑字母数字字符并忽略大小写。通过双指针从两端向中间扫描,跳过非字母数字字符进行比较。
392

被折叠的 条评论
为什么被折叠?



