[LeetCode/LintCode] Valid Palindrome

Ververica 发布于2019-08-14 15:16 / 2624人阅读

Valid Palindrome Problem

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

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.

重点是.toLowerCase()和Character.isLetterOrDigit()两个函数的使用；
以及指针在标点情况下移动后的continue语句。

Challenge

O(n) time without extra memory. ----> 指针嘛。

Solution

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return true;
        }
        s = s.toLowerCase();
        int start = 0, end = s.length() - 1;
        while (start < end) {
            if (!Character.isLetterOrDigit(s.charAt(start))) {
                start++;
                continue;
            }
            if (!Character.isLetterOrDigit(s.charAt(end))) {
                end--;
                continue;
            }
            if (s.charAt(start) != s.charAt(end)) {
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
}

Update 2018-9

class Solution {
    public boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) return true;
        s = s.trim().toLowerCase();
        int i = 0, j = s.length()-1;
        while (i <= j) {
            while (i <= j && !((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) >= "0" && s.charAt(j) <= "9"))) j--;
            if (i <= j && s.charAt(i++) != s.charAt(j--)) return false;
        }
        return true;
    }
}

GPU云服务器云服务器 palindrome Valid valid?file VALID_ARCHS

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