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[LeetCode]Valid Palindrome

时间:2014-12-03 23:34:37      阅读:396      评论:0      收藏:0      [点我收藏+]

标签:java   leetcode   

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.

双指针,一个从左往右,一个从右往左,相遇比较

public class Solution {
	public boolean isPalindrome(String s) {
		String str = s.toLowerCase();
		StringBuilder sb = new StringBuilder();
		StringBuilder rev = new StringBuilder();
		int i=0,j=s.length()-1;
		while(i<=j){
			while(i<j&&!(('a'<=str.charAt(i)&&str.charAt(i)<='z')||('0'<=str.charAt(i)&&str.charAt(i)<='9'))){
				i++;
			}
			while(i<j&&!(('a'<=str.charAt(j)&&str.charAt(j)<='z')||('0'<=str.charAt(j)&&str.charAt(j)<='9'))){
				j--;
			}
			sb.append(str.charAt(i));
			rev.append(str.charAt(j));
			i++;
			j--;
		}
		if(sb.toString().equals(rev.toString())){
			return true;
		}
		return false;
	}
}

正常遍历,提取有用的字符串,反序然后做比较。

public class Solution {
	public boolean isPalindrome(String s) {
		String str = s.toLowerCase();
		StringBuilder sb = new StringBuilder();
		StringBuilder rev = new StringBuilder();
		for(int i=0;i<str.length();i++){
			char c = str.charAt(i);
			if(('a'<=c&&c<='z')||('0'<=c&&c<='9')){
				sb.append(c);
			}
		}
		for(int i=sb.length()-1;i>=0;i--){
			rev.append(sb.charAt(i));
		}
		if(sb.toString().equals(rev.toString())){
			return true;
		}
		return false;
	}
}




[LeetCode]Valid Palindrome

标签:java   leetcode   

原文地址:http://blog.csdn.net/guorudi/article/details/41703961

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